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Mathematics  for  Common  Schools 


MANUAL   FOR   TEACHERS 


INCLUDING 


DEFINITIONS,  PRINCIPLES,  AND  RULES 

AND   SOLUTIONS   OF  THE   MORE 

DIFFICULT    PROBLEMS 


BY 


JOHN   H.   WALSH 

ASSOCIATE  SUPERINTENDENT  OF  PUBLIC   INSTRUCTIOlf 
BROOKLYN,   N.Y. 


HIGHERABIIHMETIC 

^B  ft  A  a 

CAUFOg 

BOSTON,  U.S.A. 

D.  C.  HEATH  &  CO.,  PUBLISHERS 

1898 


W3      '     . 


Copyright,  18»f^ 
Bt  JOHN  H.  WALSH. 


J.  S.  Gushing  &  Co.  —  Berwick  &  Smith. 
Boston,  Mass..  U.S.A. 


CONTENTS 

(Higher  Arithmetic  Manual.) 


1  PAGE 

Introductory 1 

Plan  and  scope  of  the  work  —  Grammar  school  algebra  —  Con- 
structive geometry. 

II 

General  Hints 5 

Division  of  the  work  —  Additions  and  omissions  —  Oral  and 
written  work  —  Use  of  books  —  Conduct  of  the  recitation  —  Drills 
and  sight  work  —  Definitions,  principles,  and  rules  —  Language 
—  Analysis  —  Obj  active  illustrations  —  Approximate  answers  — 
Indicating  operations  —  Paper  vs.  slates. 

XIII 
Notes  on  Chapter  Ten 87 

XIV 

Notes  on  Chapter  Eleven 94 

XV 

Notes  on  Chapter  Twelve 117 

XVI 

Notes  on  Chapter  Thirteen 149 

XVII 

Notes  on  Chapter  Fourteen 185 

iii 


IV  CONTENTS 

XVIII  PAG. 

Notes  on  Chaptee  Fifteen 225 

XIX 
Notes  on  Chapteb  Sixteen 239 

XX 

Notes  on  the  Appendix 282 

SUPPLEMENT. 

Definitions,  Principles,  and  Rules i 

General  definitions  —  Additions  —  Subtraction  —  Multiplication 

—  Division  —  Factoring  —  Cancellation  —  Greatest  common  divi- 
sor —  Least  common  multiple  —  Fractions  —  Decimals  —  Accounts 
and  bills  —  Denominate  numbers  —  Percentage  —  Profit  and   loss 

—  Commercial  discount  —  Commission  —  Insurance  —  Taxes  — 
Duties  —  Interest  —  Partial  payments  —  Bank  discount  —  Ex- 
change —  Equation  of  payments  —  Ratio  —  Proportion  —  Partner- 
ship —  Involution  —  Evolution  —  Stocks  and  bonds  —  Notes, 
drafts,  and  checks. 

Answebs 1 


MANUAL   FOR   TEACHERS 

I 

INTRODUCTOEY 

Plan  and  Scope  of  the  Work.  —  In  addition  to  the  subjects 
generally  included  in  the  ordinary  i^ext-books  in  arithmetic, 
Mathematics  for  Common  Schools  contains  such  simple  work 
in  algebraic  equations  and  constructive  geometry  as  can  be 
studied  to  advantage  by  pupils  of  the  elementary  schools. 

The  arithmetical  portion  is  divided  into  thirteen  chapters, 
each  of  which,  except  the  first,  contains  the  work  of  a  term  of 
five  months.  The  following  extracts  from  the  table  of  contents 
will  show  the  arrangement  of  topics  : 

First  and  Second  Years 

Chapter  I.  —  Numbers  of  Three  Figures.  Addition  and  Sub- 
traction. 

Third  Year 

Chapters  11.  and  III.  —  Numbers  of  Five  Figures.  Multipli- 
ers and  Divisors  of  One  Figure.  Addition  and  Subtraction  of 
Halves,  of  Fourths,  of  Thirds.  Multiplication  by  Mixed  Num- 
bers. Pint,  Quart,  and  Gallon ;  Ounce  and  Pound.  Roman 
Notation. 

1 


MANUAL   FOR  TEACHERS 


Fourth  Year 


Ohapters  IV.  and  V.  —  Numbers  of  Six  Figures.  Multipliers  and 
Divisors  of  Two  or  More  Figures.  Addition  and  Subtraction  of 
Easy  Fractions.  Multiplication  by  Mixed  Numbers.  Simple 
Denominate  Numbers.     Roman  Notation. 


Fifth  Year 

Ohapters  VI.  and  VII.  —  Fractions.     Decimals  of  Three  Places. 
Bills.     Denominate  Numbers.     Simple  Measurements. 


Sixth  Year 

Ohapters  VIII.  and  IX.  —  Decimals.     Bills.     Denominate  Num- 
bers.    Surfaces  and  Volumes.     Percentage  and  Interest. 


Seventh  Year 

Ohapters  XI.  and  XII.  —  Percentage  and  Interest.  Commercial 
and  Bank  Discount.  Cause  and  Effect.  Partnership.  Bonds 
and  Stocks.  Exchange.  Longitude  and  Time.  Surfaces  and 
Volumes. 

Eighth  Year 

Ohapters  XIII.  and  XIV.  —  Partial  Payments.  Equation  of 
Payments.  Annual  Interest.  Metric  System.  Evolution  and 
Involution.     Surfaces  and  Volumes. 


INTEODUCTORY  3 

While  all  of  the  above  topics  are  generally  included  in  an 
eight  years'  course,  it  may  be  considered  advisable  to  omit  some 
of  them,  and  to  take  up,  instead,  during  the  seventh  and  eighth 
years,  the  constructive  geometry  work  of  Chapter  XVI.  Among 
the  topics  that  may  be  dropped  without  injury  to  the  pupil  are 
Bonds  and  Stocks,  Exchange,  Partial  Payments,  and  Equation 
of  Payments. 

Grammar  School  Algebra.  —  Chapter  X.,  consisting  of  a  dozen 
pages,  is  devoted  to  the  subject  of  easy  equations  of  one  unknown 
quantity,  as  a  preliminary  to  the  employment  of  the  equation  in 
so  much  of  the  subsequent  work  in  arithmetic  as  is  rendered 
more  simple  by  this  mode  of  treatment.  To  tea.chers  desirous 
of  dispensing  with  rules,  sample  solutions  of  type  examples,  etc., 
the  algebraic  method  of  solving  the  so-called  "  problems  "  in  per- 
centage, interest,  discount,  etc.,  is  strongly  recommended. 

In  Chapter  XV.,  intended  chiefly  for  schools  having  a  nine 
years'  course,  the  algebraic  work  is  extended  to  cover  simple 
equations  containing  two  or  more  unknown  quantities,  and  pure 
and  affected  quadratic  equations  of  one  unknown  quantity. 

No  attempt  has  been  made  in  these  two  chapters  to  treat 
algebra  as  a  science ;  the  aim  has  been  to  make  grammar-school 
pupils  acquainted,  to  some  slight  extent,  with  the  great  instru- 
ment of  mathematical  investigation,  —  the  equation. 

Oonstructive  Geometry.  —  Progressive  teachers  will  appreciate  the 
importance  of  supplementing  the  concrete  geometrical  instruction 
now  given  in  the  drawing  and  mensuration  work.  Chapter  XVI. 
contains  a  series  of  problems  in  construction  so  arranged  as  to 
enable  pupils  to  obtain  for  themselves  a  working  knowledge  of 
all  the  most  important  facts  of  geometry.  Applications  of  the 
facts  thus  ascertained,  are  made  to  the  mensuration  of  surfaces 
and  volumes,  the  calculation  of  heights  and  distances,  etc.  No 
attempt  is  made  to  anticipate  the  work  of  the  high-school  by 
teaching  geometry  as  a  science. 


4  MANUAL   FOR  TEACHERS 

While  the  construction  problems  are  brought  together  into  a 
single  chapter  at  the  end  of  the  book,  it  is  not  intended  that 
instruction  in  geometry  should  be  delayed  until  the  preceding 
work  is  completed.  Chapter  XVI.  should  be  commenced  not  later 
than  the  seventh  year,  and  should  be  continued  throughout  the 
remainder  of  the  grammar-school  course.  For  the  earlier  years, 
suitable  exercises  in  the  mensuration  of  the  surfaces  of  triangles 
and  quadrilaterals,  and  of  the  volumes  of  right  parallelopipedons 
have  been  incorporated  with  the  arithmetic  work. 


II 

GENERAL   HINTS 

Division  of  the  "Work.  —  The  five  chapters  constituting  Part  I. 
of  Mathematics  for  Comiinon  Schools  should  be  completed  by  the 
end  of  the  fourth  school  year.  The  remaining  eight  arithmetic 
chapters  constitute  half-yearly  divisions  for  the  second  four  years 
of  school.  Chapter  I.,  with  the  additional  oral  work  needed  in 
the  case  of  young  pupils,  will  occupy  about  two  years;  the  re- 
maining four  chapters  should  not  take  more  than  half  a  year  each. 
When  the  Grube  system  is  used,  and  the  work  of  the  first  two 
years  is  exclusively  oral,  it  will  be  possible,  by  omitting  much  of 
the  easier  portions  of  the  first  two  chapters,  to  cover,  during  the 
third  year,  the  ground  contained  in  Chapters  I.,  II.,  and  III. 

Additions  and  Omissions.  —  The  teacher  should  freely  supple- 
ment the  work  of  the  text-book  when  she  finds  it  necessary  to  do 
so ;  and  she  should  not  hesitate  to  leave  a  topic  that  her  pupils 
fully  understand,  even  though  they  may  not  have  worked  all  the 
examples  given  in  connection  therewith.  A  very  large  number 
of  exercises  is  necessary  for  such  pupils  as  can  devote  a  half-year 
to  the  study  of  the  matter  furnished  in  each  chapter.  In  the 
case  of  pupils  of  greater  maturity,  it  will  be  possible  to  make 
more  rapid  progress  by  passing  to  the  next  topic  as  soon  as  the 
previous  work  is  fairly  well  understood. 

Oral  and  Written  "Work.  — The  heading  "Slate  Problems"  is 
merely  a  general  direction,  and  it  should  be  disregarded  by  the 
teacher  when  the  pupils  are  able  to  do  the  work  "  mentally." 
The  use  of  the  pencil  should  be  demanded  only  so  far  as  it  may 

5 


6  MANUAL   FOR   TEACHERS 

be  required.  It  is  a  pedagogical  mistake  to  insist  that  all  of  the 
pupils  of  a  class  should  set  down  a  number  of  figures  that  are 
not  needed  by  the  brighter  ones.  As  an  occasional  exercise,  it 
may  be  advisable  to  have  scholars  give  all  the  work  required  to 
solve  a  problem,  and  to  make  a  written  explanation  of  each  step 
in  the  solution ;  but  it  should  be  the  teacher's  aim  to  have  the 
majority  of  the  examples  done  with  as  great  rapidity  as  is  con- 
sistent with  absolute  correctness.  It  will  be  found  that,  as  a 
rule,  the  quickest  workers  are  the  most  accurate. 

Many  of  the  slate  problems  can  be  treated  by  some  classes  as 
"sight"  examples,  each  pupil  reading  the  question  for  himself 
from  the  book,  and  writing  the  answer  at  a  given  signal  without 
putting  down  any  of  the  work. 

Use  of  Books.  —  It  is  generally  recommended  that  books  be 
placed  in  pupils'  hands  as  early  as  the  third  school  year.  Since 
many  children  are  unable  at  this  stage  to  read  with  sufficient 
intelligence  to  understand  the  terms  of  a  problem,  this  work 
should  be  done  under  the  teacher's  direction,  the  latter  reading 
the  questions  while  the  pupils  follow  from  their  books.  In  later 
years,  the  problems  should  be  solved  by  the  pupils  from  the 
books  with  practically  no  assistance  whatever  from  the  teacher. 

Conduct  of  the  Eecitation.  —  Many  thoughtful  educators  consider 
it  advisable  to  divide  an  arithmetic  class  into  two  sections,  for 
some  purposes,  even  where  its  members  are  nearly  equal  in 
attainments.  The  members  of  one  division  of  such  a  class  may 
work  examples  from  their  books  while  the  others  write  the 
answers  to  oral  problems  given  by  the  teacher,  etc. 

Where  a  class  is  thus  taught  in  two  divisions,  the  members  of 
each  should  sit  in  alternate  rows,  extending  from  the  front 
of  the  room  to  the  rear.  Seated  in  this  way,  a  pupil  is  doing  a 
different  kind  of  work  from  those  on  the  right  and  the  left,  and 
he  would  not  have  the  temptation  of  a  neighbor's  slate  to  lead 
him  to  compare  answers. 


GENERAL   HINTS  7 

As  an  economy  of  time,  explanations  of  new  subjects  might  be 
given  to  the  whole  class;  but  much  of  the  arithmetic  work 
should  be  done  in  "sections,"  one  of  which  is  under  the  im- 
mediate direction  of  the  teacher,  the  other  being  employed 
in  "seat"  work.  In  the  case  of  pupils  of  the  more  advanced 
classes,  "seat"  work  should  consist  largely  of  "problems"  solved 
without  assistance.  Especial  pains  have  been  taken  to  so  grade 
the  problems  as  to  have  none  beyond  the  capacity  of  the  average 
pupil  that  is  willing  to  try  to  understand  its  terms.  It  is  not 
necessary  that  all  the  members  of  a  division  .should  work  the 
same  problems  at  a  given  time,  nor  the  same  number  of  prob- 
lems, nor  that  a  new  topic  should  be  postponed  until  all  of  the 
previous  problems  have  been  solved. 

Whenever  it  is  possible,  all  of  the  members  of  the  division 
working  under  the  teacher's  immediate  direction  should  take 
part  in  all  the  work  done.  In  mental  arithmetic,  for  instance, 
while  only  a  few  may  be  called  upon  for  explanations,  all  of  the 
pupils  should  write  the  answers  to  each  question.  The  same  is 
true  of  much  of  the  sight  work,  the  approximations,  some  of  the 
special  drills,  etc. 

Drills  and  Sight  Work.  —  To  secure  reasonable  rapidity,  it  is 
necessary  to  have  regular  systematic  drills.  They  should  be 
employed  daily,  if  possible,  in  the  earlier  years,  but  should  never 
last  longer  than  five  or  ten  minutes.  Various  kinds  are  sug- 
gested, such  as  sight  addition  drills,  in  Arts.  3,  11,  24,  26,  etc. ; 
subtraction,  in  Arts.  19,  50,  53,  etc. ;  multiplication,  in  Arts.  71, 
109,  etc. ;  division,  in  Arts.  199,  202,  etc. ;  counting  by  2's,  3's, 
etc.,  in  Art.  61 ;  carrying,  in  Art.  53,  etc.  For  the  young  pupil, 
those  are  the  most  valuable  in  which  the  figures  are  in  his  sight, 
and  in  the  position  they  occupy  in  an  example ;  see  Arts.  3,  34, 
164,  etc. 

Many  teachers  prepare  cards,  each  of  which  contains  one  of 
the  combinations  taught  in  their  respective  grades.  Showing 
one  of  these  cards,  the  teacher  requires  an  immediate  answer 


8  MANUAL   FOR  TEACHERS 

from  a  pupil.  If  his  reply  is  correct,  a  new  card  is  shown  to 
the  next  pupil,  and  so  on.  Other  teachers  write  a  number  of 
combinations  on  the  blackboard,  and  point  to  them  at  random, 
requiring  prompt  answers.  When  drills  remain  on  the  board 
for  any  considerable  time,  some  children  learn  to  know  the 
results  of  a  combination  by  its  location  on  the  board,  so  that 
frequent  changes  in  the  arrangement  of  the  drills  are,  therefore, 
advisable.  The  drills  in  Arts.  Ill,  112,  and  115  furnish  a  great 
deal  of  work  with  the  occasional  change  of  a  single  figure. 

For  the  higher  classes,  each  chapter  contains  appropriate 
drills,  which  are  subsequently  used  in  oral  problems.  It  happens 
only  too  frequently  that  as  children  go  forward  in  school  they 
lose  much  of  the  readiness  in  oral  and  written  work  they 
j)ossessed  in  the  lower  grades,  owing  to  the  neglect  of  their 
teachers  to  continue  to  require  quick,  accurate  review  work  in 
the  operations  previously  taught.  These  special  drills  follow 
the  plan  of  the  combinations  of  the  earlier  chapters,  but  gradu- 
ally grow  more  difficult.  They  should  first  be  used  as  sight 
exercises,  either  from  the  books  or  from  the  blackboard. 

To  secure  valuable  results  from  drill  exercises,  the  utmost 
possible  promptness  in  answers  should  be  insisted  upon. 

Definitions,  Principles,  and  Eules.  —Young  children  should  not 
memorize  rules  or  definitions.  They  should  learn  to  add  by 
adding,  after  being  first  shown  by  the  teacher  how  to  perform 
the  operation.  Those  not  previously  taught  by  the  Grube 
method  should  be  given  no  reason  for  "  carrying."  In  teaching 
such  children  to  write  numbers  of  two  or  three  figures,  there  is 
nothing  gained  by  discussing  the  local  value  of  the  digits.  Dur- 
ing the  earlier  years,  instruction  in  the  art  of  arithmetic  should 
be  given  with  the  least  possible  amount  of  science.  While  prin- 
ciples may  be  incidentally  brought  to  the  view  of  the  children 
at  times,  there  should  be  no  cross-examination  thereon.  It  may 
be  shown,  for  instance,  that  subtraction  is  the  reverse  of  addition, 
and  that  multiplication  is  a  short  method  of  combining  equal 


GENERAL    HINTS  "'       "  ""^  9 

numbers,  etc. ;  but  care  should  be  taken  in  the  case  of  pupils 
below  about  the  fifth  school  year  not  to  dwell  long  on  this  side 
of  the  instruction.  By  that  time,  pupils  should  be  able  to  add, 
subtract,  multiply,  and  divide  whole  numbers ;  to  add  and  sab- 
tract  simple  mixed  numbers,  and  to  use  a  mixed  number  as  a 
multiplier  or  a  multiplicand  ;  to  solve  easy  problems,  with  small 
numbers,  involving  the  foregoing  operations  and  others  contain- 
ing the  more  commonly  used  denominate  units".  Whether  or  not 
they  can  explain  the  principles  underlying  the  operations  is  of 
next  to  no  importance,  if  they  can  do  the  work  with  reasonable 
accuracy  and  rapidity. 

When  decimal  fractions  are  taken  up,  the  principles  of  Arabic 
notation  should  be  developed  ;  and  about  the  same  time,  or  some- 
what later,  the  principles  upon  which  are  founded  the  operations 
in  the  fundamental  processes,  can  be  briefly  discussed. 

Definitions  should  in  all  cases  be  made  by  the  pupils,  their 
mistakes  being  brought  out  by  the  teacher  through  appropriate 
questions,  criticisms,  etc.  Systematic  work  under  this  head 
should  be  deferred  until  at  least  the  seventh  year. 

The  use  of  unnecessary  rules  in  the  higher  grades  is  to  be 
deprecated.  When,  for  instance,  a  pupil  understands  that  per 
cent  means  hundredths,  that  seven  per  cent  means  seven  hun- 
dredths, it  should  not  be  necessary  to  tell  him  that  7  per  cent  of 
143  is  obtained  by  multiplying  143  by  .07.  It  should  be  a  fair 
assumption  that  his  previous  work  in  the  multiplication  of 
common  and  of  decimal  fractions  has  enabled  him  to  see  that 
7  per  cent  of  143  is  ^  of  143  or  143  X  .07,  without  information 
other  than  the  meaning  of  the  term  "  per  cent." 

When  a  pupil  is  able  to  calculate  that  15  %  of  120  is  18,  he 
should  be  allowed  to  try  to  work  out  for  himself,  without  a  rule, 
the  solution  of  this  problem :  18  is  what  per  cent  of  120  ?  or  of 
this:  18  is  15%  of  what  number?  These  questions  should 
present  no  more  difliculty  in  the  seventh  year  than  the  following 
examples  in  the  fifth  :  (a)  Find  the  cost  of  ^  ton  of  hay  at  $12 
per  ton.     (b)  When  hay  is  worth  $12  per  ton,  what  part  of  a 


10  MANUAL   FOR   TEACHERS 

ton  can  be  bought  for  $  1.80  ?  (c)  If  ^  ton  of  hay  costs  $1.80, 
what  is  the  value  of  a  ton  ? 

When,  however,  it  becomes  necessary  to  assist  pupils  in  the 
solution  of  problems  of  this  class,  it  is  more  profitable  to  furnish 
them  with  a  general  method  by  the  use  of  the  equation,  than 
with  any  special  plan  suited  only  to  the  type  under  immediate 
discussion. 

In  the  supplement  to  the  Manual  will  be  found  the  usual  defini- 
tions, principles,  and  rules,  for  the  teacher  to  use  in  such  a  way 
as  her  experience  shows  to  be  best  for  her  pupils.  The  rules 
given  are  based  somewhat  on  the  older  methods,  rather  than  on 
those  recommended  by  the  author.  He  would  prefer  to  omit 
entirely  those  relating  to  percentage,  interest,  and  the  like  as 
being  unnecessary,  but  that  they  are  called  for  by  many  success- 
ful teachers,  who  prefer  to  continue  the  use  of  methods  which 
they  have  found  to  produce  satisfactory  results. 

Language.  —  While  the  use'  of  correct  language  should  be 
insisted  upon  in  all  lessons,  children  should  not  be  required  in 
arithmetic  to  give  all  answers  in  "  complete  sentences."  Espe- 
cially in  the  drills,  it  is  important  that  the  results  be  expressed 
in  the  fewest  possible  words. 


—  Sparing  use  of  analyses  is  recommended  for  begin- 
ners. If  a  pupil  solves  a  problem  correctly,  the  natural  inference 
should  be  that  his  method  is  correct,  even  if  he  be  unable  to  state 
it  in  words.  When  a  pupil  gives  the  analysis  of  a  problem,  he 
should  be  permitted  to  express  himself  in  his  own  way.  Set 
forms  should  not  be  used  under  any  circumstances. 

Objective  Illustrations.  —  The  chief  reason  for  the  use  of  objects 
in  the  study  of  arithmetic  is  to  enable  pupils  to  work  without 
them.  While  counters,  weights  and  measures,  diagrams,  or  the 
like  are  necessary  at  the  beginning  of  some  topics,  it  is  important 
to  discontinue  their  use  as  soon  as  the  scholar  is  able  to  proceed 
without  their  aid. 


GENEKAL   HINTS  11 

Approximate  Answers.  —  An  important  drill  is  furnished  in 
the  "  approximations."  (See  Arts.  521,  669,  719,  etc.)  Pupils 
should  be  required  in  much  of  their  written  work  to  estimate 
the  result  before  beginning  to  solve  a  problem  with  the  pencil. 
Besides  preventing  an  absurd  answer,  this  practice  will  also  have 
the  effect  of  causing  a  pupil  to  see  what  processes  are  necessary. 
In  too  many  instances,  work  is  commenced  upon  a  problem  before 
the  conditions  are  grasped  by  the  youthful  scholar ;  which  will 
be  less  likely  to  occur  in  the  case  of  one  who  has  carefully 
"  estimated  "  the  answer.  The  pupil  will  frequently  find,  also, 
that  he  can  obtain  the  correct  result  without  using  his  pencil 
at  all. 

Indicating  Operations.  —  It  is  a  good  practice  to  require  pupils 
to  indicate  by  signs  all  of  the  processes  necessary  to  the  solution 
of  a  problem,  before  performing  any  of  the  operations.  This  fre- 
quently enables  a  scholar  to  shorten  his  work  by  cancellation,  etc. 
In  the  case  of  problems  whose  solution  requires  tedious  processes, 
some  teachers  do  not  require  their  pupils  to  do  more  than  to 
indicate  the  operations.  It  is  to  be  feared  that  much  of  the  lack 
of  facility  in  adding,  multiplying,  etc.,  found  in  the  pupils  of 
the  higher  classes  is  due  to  this  desire  to  make  work  pleasant. 
Instead  of  becoming  more  expert  in  the  fundamental  operations, 
scholars  in  their  eighth  year  frequently  add,  subtract,  multiply, 
and  divide  more  slowly  and  less  accurately  than  in  their  fourth 
year  of  school. 

Paper  vs.  Slates.  —  To  the  use  of  slates  may  be  traced  very  much 
of  the  poor  work  now  done  in  arithmetic.  A  child  that  finds  the 
sum  of  two  or  more  numbers  by  drawing  on  his  slate  the  number 
of  strokes  represented  by  each,  and  then  counting  the  total,  will 
have  to  adopt  some  other  method  if  his  work  is  done  on  material 
that  does  not  permit  the  easy  obliteration  of  the  tell-tale  marks. 
When  the  teacher  has  an  opportunity  to  see  the  number  of 
attempts  made  by  some  of  her  pupils  to  obtain  the  correct  quo- 


12  MANUAL   FOR  TEACHERS 

tient  figures  in  a  long  division  example,  she  may  realize  the 
importance  of  such  drills  as  will  enable  them  to  arrive  more 
readily  at  the  correct  result. 

The  unnecessary  work  now  done  by  many  pupils  will  be  very 
much  lessened  if  they  find  themselves  compelled  to  dispense  with 
the  "  rubbing  out "  they  have  an  opportunity  to  indulge  in  when 
slates  are  employed.  The  additional  expense  caused  by  the 
introduction  of  paper  will  almost  inevitably  lead  to  better  results 
in  arithmetic.  The  arrangement  of  the  work  will  be  looked 
after ;  pupils  will  not  be  required,  nor  will  they  be  permitted,  to 
waste  material  in  writing  out  the  operations  that  can  be  per- 
formed mentally ;  the  least  common  denominator  will  be  deter- 
mined by  inspection  ;  problems  will  be  shortened  by  the  greater 
use  of  cancellation,  etc.,  etc.  Better  writing  of  figures  and  neater 
arrangement  of  problems  will  be  likely  to  accompany  the  use  of 
material  that  will  be  kept  by  the  teacher  for  the  inspection  of 
the  school  authorities.  The  endless  writing  of  tables  and  the 
long,  tedious  examples  now  given  to  keep  troublesome  pupils 
from  bothering  a  teacher  that  wishes  to  write  up  her  '•ecords, 
will,  to  some  extent,  be  discontinued  when  slates  are  n-*  longer 
used. 


XIII 

NOTES  ON  CHAPTER  TEN 

The  formal  study  of  algebra  belongs  to  the  high-school ;  but 
some  so-called  arithmetical  problems  are  so  much  simplified  by 
the  use  of  the  equation  that  it  is  a  mistake  for  a  teacher  not  to 
avail  herself  of  this  means  of  lightening  her  pupils'  burdens. 

In  beginning  this  part  of  her  mathematical  instruction,  the 
teacher  should  not  bewilder  her  scholars  with  definitions.  The 
necessary  terms  should  be  employed  as  occasion  requires,  and 
without  any  explanation  beyond  that  which  is  absolutely  neces- 
sary. 

849.  Very  young  pupils  can  give  answers  to  most  of  these 
questions  ;  so  that  there  will  be  no  need,  for  the  present,  at  least, 
of  introducing  a  number  of  axioms  to  enable  the  scholar  to 
obtain  a  result  that  he  can  reach  without  them. 

850.  Pupils  will  learn  how  to  work  these  problems  by  work- 
ing a  number  of  them.  They  may  need  to  be  told  that  x  stands 
for  1 X ;  and  that,  as  a  rule,  only  abstract  numbers  are  used  in 
the  equations,  the  denomination  —  dollars,  marbles,  etc.  —  being 
supplied  afterwards. 

While  the  scholars  should  be  required  to  furnish  rather  full 
solutions  of  the  earlier  problems,  they  should  be  permitted  to 
shorten  the  work  by  degrees,  writing  only  whatever  may  be 
necessary. 

4.  a:-|-2a:  =  54.  8.    a: -f 2 a: 4- 6 ar - 27000. 

5.  x-\-bx=lS.  9.    x-\-bx=-12. 

6.  7a;-f  5.r=156.  10.    rr -f  2 a: -f- 3 a;  =  54. 

7.  ^x~^x  =  m.  XI.    a:-}-6^  =  42. 

87 


88  MANUAL   FOR   TEACHERS 

12.  2a;  +  10a:  =  96. 

13.  Let  X  —  the  fourth ;  then  4  a;  =  the  third,  12  a;  =  the  second, 
and  24  a;  =  the  first. 

a:  +  4a:-f-12a:4-24a:  =  41. 

14.  X  =the  second,  2  a;  =  first,  9  a:  =  third. 

15.  5a:  +  4a;  =  81.  17.    4a;  =  340. 

16.  24  a;  =  456.  19.    3  a; +  4  a:  =175. 

20.  Let  X  =  each  boy's  share  ;  2  a;  =  each  girl's  share. 

2a;  +  4a;  =  240. 

21.  a;  =  number   of  days    son  worked;  2 a:  =  number   father 
worked.     3  a;  =  son's  earnings  ;  8  a:  =  father's  earnings. 

3a;4-8a;=165. 

22.  a;  =  number  of  dimes;   2 a:  =  number   of  nickels;   6a:  = 
number  of  cents. 

(10xa;)  +  (5x2a;)+(lx6a:)  =  78, 
or  10a;  +  10a:4-6a;  =  78. 

23.  15a;— 12a;  =  75. 

24.  a;  +  4a;  +  a:  +  4a;  =  250. 

25.  Let  X  =  cost  of  speller ;  then  3  a;  =  cost  of  reader. 

26.  Let  a;  =  smaller ;  then  5a:  =  larger. 

27.  Let  a:  =  Susan's  number ;  2 a;  =  Mary's;  3 a:  =  Jane's. 

851.  10  :  1  a;  is  the  same  as  -• 

^  3 

852.  Pupils  already  know  that  J  means  3  h-  4,  so  that  they 

can  understand  that  —  means  3  a: -h- 4,  or  4  of  3a:.     When  4-  of 
4 

something  (3  a:)  is  24,  the  whole  thing  (3  a:)  must  be  4  times  24, 

|a 

4 


or  96  ;  that  is.  when  ^  =  24,  3  a:  =  96. 


When  ^  =  24,  2y  =  24  X 3,  or  72. 

o 

When  ~  =  20,  4  2  =  20  X  5,  or  100. 

5 

From  these  examples  can  be  formulated  the  rule  for  disposing 

of  a  fraction  in  one  term  of  an  equation,  which  is,  to  multiply 


NOTES   ON   CHAPTER   TEN  89 

both  terms  by  the  denominator  of  the  fraction.     In  changing  the 

3  a; 
first  term  of  the  equation,  —  =  24,  to  3  a;,  it  has  been  multi- 
plied by  4,  so  that  the  second  term  must  also  be  multiplied  by  4. 

853.  In  solving  these  examples  by  the  algebraic  method  of 
"  clearing  of  fractions,"  attention  may  be  called  to  its  similarity 
to  the  arithmetical  method.  To  find  the  value  of  y  in  2,  the 
pupil  multiplies  8  by  5  and  divides  the  product  by  2 ;  as  an  ex- 
ample in  arithmetic,  he  would  divide  8  by  ■§-,  that  is,  he  would 
multiply  8  by  -|;  the  only  difference  being  that  by  the  latter 
method  he  would  cancel. 

While  -r^  =  8  may  be  changed  to  ^  =  4  by  dividing  both  terms 
o  5 

by  2,  beginners  are  usually  advised  to  begin  by  '*  clearing  of 
fractions,"  short  methods  being  deferred  to  a  later  stage. 

854.  6  may  be  written  ^  -j-  5£  =  92. 

8.    2|-a;  should  be  reduced  to  an  improper  fraction,  making  the 

23a; 
equation,  -— -  =  115.     Make  similar  changes  in  12,  14, 18,  and  20. 

8 

855.  2.   a; -1-^  =  100. 

^-   i  +  f  =  ?'2a:  +  a:  =  267. 

6.      ?^-?^=:15. 


9.  Let  5 a;  =  numerator  \lx  =  denominator.  7 a;  —  5  a;  =  24 ; 
2  a;  =  24 ;  a;  —  12.  The  numerator,  5  a;,  will  be  5  times  12,  or  60 ; 
the  denominator  will  be  84 ;  and  the  fraction,  fj.  Arts. 

10.    Let  X  —  greater ;  -  =  less. 

a;  +  ^=:480. 


90  MANUAL   FOR  TEACHERS 

Clearing  of  fractions,         7x-\-x  =  3360, 

8a;  =  3360, 
X  =  420,  the  greater  number, 

^  =  60,  the  less. 

Or,  ,  let  a:  =  less ;  7x  =  greater, 

a:  +  7a:  =  480, 
8  a:  =  480, 

X  =  60,  the  less, 
*lx  =  420,  the  greater. 

The  employment  of  the  latter  plan  does  away  with  fractions 
ib  the  original  equation. 

11.    30a:-a:  =  522,  ora:-;^=522. 

30 

13.    Let   a:  =  number   of    plums;    4a:  =  number    of  peaches. 

Then  2a:  will  be  cost  of  plums,  and  12a:  the  cost  of  the  peaches. 

2a:+12a:  =  70. 

16.  a:-?^=:80. 

7 

17.  a:-?^-?  =  24. 

8       4 

18.  X  +  l^a;  -j-(^xx^)  =  15. 
a:  +  ^  +  5a:  =  15. 

19.  Let  X  =  price  per  yard  of  the  48-yard  piece  ;  2a:  =  price 
per  yard  of  the  36-yard  piece ;  48a:  will  be  the  total  cost  of  one, 
and  72  a:,  of  the  other. 

48a: -I- 72a:  =  240. 

20.  160  a: -f  120  a:  =  840. 

856.  The  pupils  should  be  permitted  to  give  these  answers 
without  assistance. 

In  Art.  857  is  explained  what  is  meant  by  "  transposing." 


2a;- 

-  «=i«+|-| 

.2a;- 

-36  =  96  +  3a;-2a; 

2x- 

-3a; 4-2a;=:  96  +  36 

11a;- 132 

a;  =  12 

NOTES   ON  CHAPTER  TEN  91 

858.  While  these  exercises  are  so  simple  that  they  can  be 
worked  without  a  pencil,  they  should  be  used  to  show  the  steps 
generally  taken  in  more  complicated  equations.        ,  07  _  e^ 

In  1,  for  instance,  the  work  should  take  the  _ 

form  here  indicated,  only  a  single  step  being  _ 

taken  at  a  time.     In  19,  the  first  step  is  to 

clear  the  equation  of  fractions  by  multiplying  by  6 ;  the  second 

step  is  to  transpose  the  unknown 
quantities  to  the  left  side  of  the 
equation,  and  the  known  quantities 
to  the  right ;  the  third  step  is  to  com- 
bine the  unknown  quantities  into 
one,  and  to  make  a  similar  combina- 
tion  of   the   known   quantities;  the 

last  step  is  to  find  the  value  of  x. 

After  a  little  more  familiarity  with  exercises  of  this  kind,  the 

pupil  can  take  short  cuts  with  less  danger  of  mistakes ;  for  the 

present,  however,  it  will  be  safer  to  proceed  in  the  slower  way. 

859.  5.   a;  +  (a;+75)4-a:-f(a;  +  75)  =  250. 

a;  +  a;  -f-  a;  4-  a;  =  250  -  75  —  75. 

Note.  —  The  parentheses  used  here  are  unnecessary.  They  are  employed 
merely  to  show  that  a;  +  75  is  one  side  of  the  field. 

6.  a;  +  (a;  +  8)  =  86.  9.   a;  +  a;  +  72  =  96. 

7.  a;  +  a; +  318  =  2436.  10.   a;-f-f  =45. 

o      4 

8.   a;+  1+7  =  100. 

11.  X  =  one  part ;  2a;  —  6  =  other  part. 

a;  +  2a;-6  =  45. 

12.  a;  =  John's  money  ;  a;  +  5  =  William's  money. 

3a;+15  +  5a;=103. 


92  MANUAL   FOR   TEACHERS 

13.  Let  X  =  price  of  a  horse ;         a;  —  80  =  price  of  a  cow ; 
4a;  =  cost  of  four  horses ;    Sx—  240  =  cost  of  three  cows. 

4a;  4- 3a;    -240  =  635, 
7  a;  =  635 +  240  =  875, 
X  =  125,  price,  in  dollars,  of  a  horse  ; 
a;  —  80  =    45,  price,  in  dollars,  of  a  cow. 
Other  pupils  may  solve  the  problems  in  this  way : 

X  =  price  of  a  cow ;  a;  -j-  80  =  price  of  a  horse. 
3a;+  4a;  +  320  =  635, 
7  a;  =  635 -320  =  315, 
X  =    45,  price,  in  dollars,  of  a  cow  ; 
a;  +  80  =  125,  price,  in  dollars,  of  a  horse. 

14.  a;  =  number  of    dimes;    a; -f- H  =  number   of    five-cent 

pieces;    10a;  =  value  of  dimes  (in  cents);    5 a; -}- 55  =  value  of 

five-cent  pieces. 

10  a;  +  5a; +  55  =  100. 

15.  a;  =  greater  ;  a;  —  48  =  less. 

a;  +  a;-48  =  100. 
Or,  X  =  less ;  a;  +  48  =  greater. 

a;  +  a;  +  48  =  100. 

17.  a;  =  share  of  the  first ; 

X  +  2400  =  share  of  the  second  ; 

X  +  2400  +  2400  =  share  of  the  third. 

a;  +  a;  +  2400  +  a;  +  2400  +  2400  =  18000. 

18.  Let  X  =  less ;  a;  +  33  =  greater. 

a; +  33 -3a:  =11. 
Bringing  known  quantities  to  the  left  side  of  the  equation,  and 
the  unknown  quantities  to  the  right, 

33-ll  =  3a;-ar, 
22  =  2a;, 
11  =  a;. 


( -CrNiTEKSITT  I 

NOTES   ON   CHAPTER   tt^Wl^CA U fOT^>^  93 

Or,  a; -3a;  =11  — 33, 

-2a;  =  -22. 
Changing  signs  of  both  terms, 

2rr=22, 
a;  =11. 
This  problem  may  also  be  worked  in  this  way  : 
X  —  less  ;  3  a:  +  11  =  greater. 
3a;+ll-a;=33. 

19.  X  =  number  of  5-cent  stamps  ;  a;  +  15  =  number  of  2-cent 
stamps  ;  x  -}- SO  =  number  of  postal  cards. 

5  a;  +  2  a;  +  30  +  a:  +  30  =  100. 

20.  a;  =  number  of  horses;  a; +  17  =  number  of  cows;  2a;-f-39 
=  number  of  sheep. 

a;  +  a;+17  +  2a;  +  39  =  88. 


XIV 
NOTES  ON  CHAPTER  ELEVEN 

With  this  chapter  begins  the  regular  work  in  percentage,  and 
it  is  important  that  the  pupils  obtain,  as  soon  as  possible,  a 
correct  idea  of  what  is  meant  by  the  term  per  cent. 

Many  of  the  various  subdivisions  of  this  topic  found  in  some 
books,  are  taken  up  only  incidentally,  while  others  are  omitted 
altogether,  the  aim  being  to  give  the  pupils  a  foundation  upon 
which  they  can  subsequently  build,  rather  than  to  scatter  their 
energies  over  too  great  a  diversity  of  subjects. 

860.  The  reduction  of  a  common  fraction  to  a  per  cent,  con- 
sists in  changing  the  former  to  a  decimal  of  two  places.  In  re- 
ducing ^  to  a  decimal,  the  result  is  .5,  or  5  tenths ;  in  changing 
■J-  to  an  equivalent  per  cent,  the  result  is  50  per  cent,  or  50 
hundredths.  In  reducing  ^  to  a  decimal,  the  answer  is  given  in 
three  places,  .125,  or  125  thousandths ;  in  changing  it  to  a  per 
cent,  the  division  is  stopped  at  the  second  place,  and  the  remain- 
der written  as  a  fraction,  12^  per  cent,  or  12^  hundredths. 

The  denominator  of  a  per  cent  being  always  the  same,  100, 
the  comparative  value  of  several  per  cents  is  known  at  sight. 
To  compare  f  and  f  as  common  fractions,  they  must  be  changed 
to  }4  and  J^;  if  a  further  comparison  is  to  be  made  between 
these  and  -f^,  a  new  common  denominator  must  be  employed, 
and  the  fractions  reduced  to  -^^^  -^^  and  ■^.  Changing  the 
fractions  to  decimals,  625  thousandths,  6  tenths,  and  58^^  hun- 
dredths, simplifies  the  comparison  ;  but  it  is  still  easier  to  de- 
termine their  relative  value  when  they  are  expressed  as  62J  per 
cent,  60  per  cent,  and  58J  per  cent. 


NOTES  ON  CHAPTER  ELEVEN  95 

The  teacher  must  not  be  discouraged  if  the  pupil  fails  to  grasp 
at  once  the  full  meaning  of  percentage,  Definitions  will  not 
help  materially  ;  much  practice  in  working  examples  is  necessary 
to  give  the  knowledge  desired. 

863.  Many  children  find  it  diflicult  to  distinguish  between 
^%  and  50%.  If  the  former  is  read  in  the  business  way,  -J-  of 
one  per  cent,  it  may  make  the  distinction  plainer. 

864.  Per  cents  being  generally  given  in  two  figures,  scholars 
hesitate  to  give  the  correct  answers:  300%,  250%,  125%, 
16331%,  420%,  910%. 

865.  While  pupils  will  find  33|%  of  81  cows,  by  dividing  81 
by  3,  they  should  understand  that  they  are  really  multiplying 
81  by  33^  hundredths,  or  81  by  |.  In  4,  6%  of  150,  or  ^^  of 
150,  may  be  obtained  by  multiplying  150  by  6  and  cutting  off 
two  ciphers  ;  or  by  dividing  150  by  100,  obtaining  1^,  and  multi- 
plying this  quotient  by  6  ;  or  by  reducing  6%  to  -^,  and  finding 
-^  of  150.  In  9,  the  pupil  should  find  1%  of  $  640  and  take  one- 
half  of  the  result. 

The  scholars  should  be  permitted  to  use  their  own  method  of 
solving  these  problems,  the  different  analyses  given  by  the  pupils 
furnishing  their  class-mates  an  opportunity  to  select  a  simpler 
method. 


Although  every  pupil  may  not  be  able  to  determine  at 
once  the  shortest  way  of  calculating  a  given  example,  no  one 
should  be  allowed  to  work  3,  by  multiplying  by  33^.  "When  the 
multiplication  by  ^  has  been  performed,  the  answer  has  been  ob- 
tained, except  as  to  the  location  of  the  decimal  point,  and  the 
waste  of  time  in  multiplying  by  3,  repeating  this  product,  and 
adding  three  columns  should  not  be  tolerated.  No  fault  should 
be  found  with  the  average  pupil  for  failing  to  recognize  in  1,  that 

^i%  ^s  ttJ    ^^  *^^^  ^^  ^^'  ^i%  i^  "TO"-     ^^^^  general  method 
should  be  to  multiply  by  the  figures  given  to  represent  the  per 


96  MANUAL    FOR   TEACHERS 

cent,  except  in  such  cases  as  12J%,  16|%,  25%,  33J%,  37J%, 
50%,  and  possibly  a  few  others. 

^      .  Where  the  given  numbers  are  used,  they  should  be 

made  the  multipliers  and  expressed  as  hun-    (uqq  a^. 

'. — ~  dredths.     Nothing  is  gained  in  5,  by  reducing         ^\,, 

<l>i.oD     ^jjg  -J-  to  a  decimal;  although  in  13,  writing    '. — I 

5^%,  .055,  might  make  it  easier  for  some. 

868.  The  rule  generally  given  of  finding  the  percentage,  by 
multiplying  by  the  rate  expressed  as  hundredths,  is  here  modified 
to  the  extent  of  using  the  common  fraction  to  express  hundredths, 
instead  of  the  decimal,  as  being  more  in  conformity  with  early 
algebraic  methods. 

The  teacher  that  prefers  to  ascertain  the  base  or  the  rate  by 
the  older  arithmetical  method,  will  omit  30-41. 

30.  _^x65  =  ^.  Am.         35.    a:-f-f=132;  etc. 
100  20  5 

31.  15^  =  26;  etc.  37.    a;-f=78;  etc. 

20  3  ' 

32.  lof  a;  =  ~  Am.  38.    -^of  ?  =  —  Am. 
4              4  100       3      150 

33.  -  =  42;  etc.  39.   -£-=^5;  etc. 
4  150       3 


1       .  X 

41.   -^  =  23;  etc. 


34.    .  +  l-Am.  40.   _of.  =  ^.  ^n.. 


800 

42.    Let  X  =  rate.     Then  -^  of  65  =  26,  or  -^  X  65  =  26. 

100  100 

In  an  equation  containing  quantities  to  be  multiplied,  the  mul- 
tiplication should  be  performed   before  the  equation  is  cleared 

65ar  \Sx 

of  fractions.     This   equation   becomes  — —  =  26,  or  -— -  =  26, 

luu  ^0 

it  being  immaterial  whether  canceling  be  done  or  not. 


NOTES  ON  CHAPTER  ELEVEN  97 

43.  After  a  little  experience  with  this  class  of  examples,  the 
equation  may  be  written  at  once,  in  the  order  in  which  the  terms 
are  given  : 

24  =  -l|-of  a;,ori|^  =  24. 
100         '       100 

44.  ?^of  a:  =  180,  or^-180. 
100  '        2 

45.  a:  +  ?  =  85. 

4 

.-3  X         n  4:  X  3 

46.  —  = of  -,  or =  — 

5      100       5'       125      5 

While  the  algebraic  method  is  of  no  advantage  to  the  bright 
scholar,  it  makes  the  employment  of  rules  unnecessary  in  the 
case  of  the  ordinary  pupil. 

48.    xX-^Xl=^-  49.    ii^  =  44. 

100  200  200 

50.  fx|.  51.   f|f  =  33. 

52.  i  of  800  =  100 ;  i%  of  800  -  1. 

63.  $175 +  i  of  $175. 

64.  2i%  of  a;  -  12.50  ;  that  is,  ^  =  12.50. 
56.  6i  X  .16. 

56.  8i  =  -^of  ?;  thatis,-^  =  ^. 

^100       3  150       3 

57.  -^  of  389.50  =  124.64  ;  §§M2^=  124.64 ; 
100  100 

389.5a;  =12464;  3895  a;  =  124640. 

58.  ?5£=  174.04 ;  95  a;  =  17404. 
100 

59.  ^  +  f|=1276. 


98  MANUAL   FOR  TEACHERS 

60.    984  =  ^  of  X ;  that  is,  ~  =  984. 
100  '  3 

62.   ^=386.75. 
4 

65.   X  =  cost  of  oats  ;  x  -\-  — —  =  1071. 

Divide  the  cost  of  the  oats  by  30^  to  find  the  number  of 
bushels. 

68.  Assessed  value  =  |  of  $  48000  =  $  32000.  Taxes  on 
32  thousand  dollars  =  $18.50  X  32. 

869.  In  giving  answers  to  these  and  to  all  other  exercises,  no 
"  guessing  "  should  be  allowed.  The  pupil  should  be  permitted 
to  obtain  the  correct  result  in  his  own  way  —  that  is,  no  inflexible 
rule  should  be  given  him  to  follow  —  but  he  should  be  able  to  get 
the  answer,  using  the  algebraic  method  if  that  seems  to  him  the 
easiest,  as  it  may  be  in  some  instances. 

The  examples  are  not  arranged  by  "  cases,"  so  that  each  will 
have  to  be  understood  before  it  can  be  worked. 

The  careless  pupil  will  probably  give  the  wrong  answer  to  13 ; 
saying  6,  instead  of  600 ;  he  will  be  likely  too,  in  14,  to  use  the 
larger  number  as  a  divisor,  and  to  obtain  44f  %  instead  of  225%. 
These  mistakes  are  less  likely  to  occur  if  he  uses  equations  — 

3  =  — —  and  — -  =  20i.     Even  those  scholars  that  have  solved 
200         100  * 

in  their  arithmetic  work  of  the  lower  grades,  examples  similar  to 
16,  will  have  new  light  thrown  on  their  method  by  using  the 

equation,  x  -{-  -  =  20.     In  mental  work,  however,  the  first  term 

5a; 
should  be  made  — — ,  to  reduce  it  in  size,  so  that  it  can  be  more 
4 

easily  remembered.  24  is  simplified  by  changing  the  fractions 
to  whole  numbers  —  ^  is  what  per  cent  of  fj,  9  is  what  per  cent 
of  10  —  before  beginning  to  calculate  the  rate.  In  25,  H  and 
6|  become  f  and  ^,  f  and  ^,  9  and  40. 

870.  1-5  can  be  worked  by  the  pupils  without  any  explana- 
tion ;  6-20  present  more  difficulty.     The   beginner  in   algebra 


NOTES  ON  CHAPTER  ELEVEN  99 

desires  to  start  at  once  with  his  x,  without  any  preliminary  cal- 
culations; and  the  usual  method  of  treating  these  examples 
requires  him  first  to  ascertain  the  gain  or  the  loss  before  com- 
mencing his  equation.  The  formula  employed  in  the  first  five 
examples  is : 

Cost  X  — —  =  gain  or  loss. 

When  the  pupil  knows  any  two  of  these  three  terms,  he  can 
calculate  the  third ;  and  6-15  furnish  data  from  which  the 
necessary  two  items  can  be  obtained.  The  pupil  must,  however, 
be  careful  in  11,  for  instance,  not  to  subtract  the  loss  from  the 
selling  price  to  obtain  the  cost. 

In  the  following  equations,  cost  X  —--  is  made  equal  to  the 
gain  or  the  loss.     No  canceling  has  been  done. 

6.    ?50^==18;6.  =  18.  7.    ^^^  ==  m. 

100  100 

8.  §^20^^  12.93,  or  ?5?^=  1293. 

100  100 

9.  ?5^i^=  181.68. 

100 

10.   §i^=5.25;    84  a:  =  525. 
100 

11.   §i£=5.25.  12.   ^^^-25. 

100  100 

13.    ^^  =  43.75;  875^  =  4375. 
100 

14.    934.56^^^^  ^^     1012.500:^^^3^^^^ 

100  100 

In  16-20,  the  cost  is  represented  by  x. 

16.  a: +  ^  =  468.75.  18.    a: +  -=  1646.08. 

4  3 

17.  a: -|=  73.84.  19.    rr-i^=204. 

u  lUU 


20. 

100 

21. 

Gain  =  i  of  $275. 

22. 

x%  of  60  =  15. 

23. 

:r  +  f=960. 
5 

100  MANUAL   FOR  TEACHERS 

27.  Saj^  +  iof  33^^. 

28.  a:-i^=33.60. 
100 

29.  Gain  =  2J%  of  $8760. 

30.  x-  — ---  — =  70. 
10     4      10 

24.  rr%of32  =  16.  31.    6000  =  a: %  of  16000. 

25.  x%  of  175  -  25.  32.   6000  =  x%  of  24000. 

26.  x%  of  200  =  25.  33.    1600  +  2J%  of  1600. 

34.   4200-3^%  of  4200. 

871.  In  1,  the  30  cu.  yd.  are  reduced  to  cubic  feet  by  multi- 
plying by  27.  Instead  of  performing  the  different  multiplications, 
they  are  merely  indicated,  so  that  work  may  be  saved  by  can- 
celing. 

Although  2  should  be  a  simple  problem  for  a  bright  pupil,  it 
is  apt  to  prove  puzzling  unless  an  x  is  introduced.  A  paste- 
board box  may  be  used  to  represent  the  walls  and  the  ceiling  of 
a  room,  the  sides  and  the  top  being  then  opened  out  to  permit 
of  its  representation  on  the  blackboard. 

3.  The  area  in  square  feet  =  ^  of  132  X  110.  This  is  reduced 
to  acres  by  dividing  by  9  X  30 J  X  160. 

132x110x4     _  . 
2  X  9  X  121  X  160 

4.  Number  of  strips  ==  6  yd.  h-  27  in.  =  6  yd.  -^  f  yd.  =  6  X  f . 

7.  The  "  development "  of  the  fence  will  be  represented  by 
four  adjoining  rectangles,  each  marked  6  ft.  high,  the  lengths 
being  25  ft.,  100  ft.,  25  ft.,  and  100  ft,  respectively,  the  whole 
forming  a  rectangle  6  ft.  X  250  ft. 

8.  A  board's  area  in  square  feet  =  12x^  =  6.  Dividing 
number  of  square  feet  in  the  fence  by  6,  gives  the  number  of 
boards. 


NOTES  ON  CHAPTER  ELEVEN  101 

9.  The  cost  of  a  square  foot  is  obtained  by  dividing  $181.50 
by  (160  X  30^  X  9)  ;  this,  multiplied  by  (300  X  200),  gives  the 
cost  of  the  plot. 

$181.50x4x300x200 
160  X  121  X  9 

The  amount  received  for  the  lots  will  be  $160  X  6. 

10.  Number  of  cakes  =  (320  X  160)  ^  (4  X  2). 

11.  Number  of  cubic  feet  ==  320  X  160  X  l^-. 

12.  (320  X  160  X  1^)  ^  (15  X  32). 

13.  Number  of  square  feet  originally  =  640  X  440.  For  build- 
ing purposes,  there  will  be  four  pieces,  each  measuring  300  ft. 
by  200  ft. 

14.  The  difference  between  the  above  areas  will  represent  the 
number  of  square  feet  in  the  streets. 

872.  Many  of  these  exercises  can  be  used  for  mental  and  sight 
work.     For  methods  of  solution,  see  Art.  870. 

873.  As  a  preliminary  to  the  formal  study  of  interest,  the 
teacher  will  need  to  see  that  her  pupils  understand  what  is 
meant  by  the  term.  She  can  explain  that  a  person  borrowing 
money  should  pay  for  its  use,  just  as  a  person  who  rents  a 
house,  etc. 

874.  In  changing  4  mo.  10  da.  to  the  fraction  of  a  year,  many 
teachers  prefer  to  reduce  the  time  to  days  and  to  write  the 
result  over  360,  -^f^,  leaving  the  reduction  to  lowest  terms  for 
the  subsequent  cancellation.  In  the  same  way,  1  yr.  5  mo.  15  da. 
is  changed  to  (360  +  150  +  15)  da.,  or  525  da.  =  f||-  yr.  The 
reduction  to  days  is  done  very  rapidly. 

875.  1.   $750xTf^x|.  5.   $360x^^X3^^^. 

2.  $84.75  x-r^X  it.  6.   $  94.43  x  yj^  X  #^. 

3.  $308.25  Xy^X^V  7.    $400x^X|fi. 

4.  $  464.75  X  yf 7  X  Iff.  etc.,  etc. 


102  MANUAL   FOR  TEACHERS 

877.  The  teacher  should  explain  that  a  person  that  owes 
money,  frequently  gives  a  note  as  an  acknowledgment  of  the 
debt,  etc. 

878.  There  is  no  general  method  applicable  to  these  problems. 

1.  Interest  for  a  year  is  $12,  or  $1  per  month,  which  gives 
5?  19  for  1  yr.  7  mo. 

2.  $3.60  per  year  is  1^  per  day,  33^  for  33  da. 

3.  $  6  per  year,  or  $  9  for  1^  yr. 

4.  $  6  per  year,  or  $  15  in  2  yr.  6  mo. 

5.  If  $  50  produces  $6  in  2  yr.,  it  will  produce  $3  in  1  yr. ; 
rate,  therefore,  is  6%. 

6.  $  18  per  year  is  $1  for  20  da.,  or  Jg-  yr. 

8.  4%  per  year-- 1%  for  90  da. ;   1%  of  $150  =  $1.50. 

9.  5%  per  year  =  |%  for  36  da. ;  \%  of  $240  =  ^  of  $2.40. 

11.  $1  is  100%  of  $1  ;  at  5%  per  year  it  will  take  20  yr. 
to  make  100%. 

12.  At  6%  it  will  take  16|yr.,  or  16  yr.  8  mo.,  to  make  100%.  ' 

13.  Disregarding  $14.90,  it  will  take  25  yr.  at  4%  to 
make  100%. 

14.  \%  per  month  =  8%  for  16  mo. ;  8%  of  $90  =  $  7.20. 

15.  5%  for  360  da.  =  1%  for  72  da. 

16.  360  da. -^  4^  =  720  da.  ^  9  =>  80  da. 

17.  5%  for  1  yr.  =  1%  for  72  da. ;  1%  of  $  75  -=  75  cents.       . 

18.  l%of$63.  20.   ^%  of  $840.  22.    1%  of  $275. 

19.  l%of$570.         21.    1%  of $150.  23.    2%  of $360. 

879.  1.  30  rd.  5  yd.  1  ft.  =  511  ft. ;  8  rd.  4  yd.  2  ft.  =  146 
ft.;  iH  =  f  ^^• 

2.    Number  of  feet  deep  =  (36  X  5)  h-  (6  X  4). 


NOTES  ON   CHAPTER   ELEVEN  103 

3.  3  mi.  96  rd.  =  1056  rd. ;  3  hr.  16  min.  =  Sj\  hr. ;  1056 
rd.  X  3^  =  1056  rd.  X  f  f  -  lif^^  rd.  =  3449f  rd.  =  10  mi.  249f 
rd.  Ans. 

4.  (1  + J)  X  (f  X  ^)  ^  (f  X  ^)  =  ^  X  f  X  ^  X  4  X  /j 

8.    The  first  two  figures  express  1800 ;  the  second  two,  5.4. 
22.   $48.37 -^8f 

880.  2.  Provisions  that  will  supply  450  men  for  5  months 
will  supply  5  times  450  men  for  1  month,  and  will  supply  (5 
times  450  men)  h-  9  for  9  months,  or  250  men.  The  number  that 
must  be  discharged  =  450  men  —  250  men  —  200  men.  Ans. 

15.  ^-^-1^=19500. 

100      100 

16.  D  bought  J  X  f  X  I  X  f  of  the  ship. 

18.  100  a; +  50^  =  340x75. 

19.  X  =  number  distributed  by  each  new  man  ]  2x  =  number 
distributed  by  each  experienced  man. 

16a;  +  32a;  =  36000. 

20.  a;  ~?=  1972.65. 

4 

881.  10.    100  cents -^1.13. 

883.   3.   $1.10+15%  of  $1.10. 

4.  9876  _  3^  J  45 

9876  =  87.2;  +  45. 

5.  640  is  what  per  cent  of  (640  +  560),  etc. 

6.  43  gal.  3  qt.  1  pt.  =  43|-  gal. ;  $70.20 -h-43|-  =  Ans. 

7.  (48  X  32) -^  (16  X  i). 

8.  20  is  what  per  cent  of  160?  20  is  what  per  cent  of  180? 

10.    Selling  price  per  bbl.  =  ^  =  ?^ ;  44%  =  —  =  — • 
^^        ^  600       4       ^'       600     24 

X       23 

Let  X  =  cost  per  bbl.    x =  — 

^  24      4 


104  MANUAL   FOR  TEACHERS 

U.   9:075^  =.24.2. 
100 

884.  425  +  99  is  1  less  than  425  +  100 ;  425  +  999  is  1  less 
than  425  + 1000. 

885.  565-99  is  1  more  than  565-100;  1424-999  is  1 
more  than  1424  -  1000. 

886.  24  X  21  =  (24  x  20)  +  (24  x  1).     See  Art.  786. 

887.  16-^  .25  =  16  H-  J  =  16  X  4  ;   36  ^  .75  =  36  ^  J  =  36 

Xj  =  12x4. 

888.  7i^|  =  ^^|  =  \<i-f-|  =  30-^3.  When  the  dividend 
is  a  whole  number,  it  is  frequently  better  to  perform  the  division 
in  the  ordinary  way  :  63  ^  3^  =  63  -^-J  =  63  X  f  =  9  X  2. 

889.  1.  $|X48.  2.  (48xf)sq.  yd.  3.  (48^f)  yd.  7. 
9  into  83i,  9  times  and  2^,  or  |,  remaining ;  9  into  9  quarters,  J. 
Ans.  9i.  10.  $lixl9.  11.  $ljxl20.  12.  (120  ^  IJ) 
yd.  =  (120  X  j\)  yd.  =  (8x8)  yd.  14.  Dimensions  of  field 
=- 80  rods  X  80  rods.  16.  95h- 4f -=  95 -^  J^  =  95  X  ^^  =  5  X  4. 
17.  4000 -f- 2.  19.  3  T.  will  cost  $15;  480  lb.@i^  per  lb. 
will  cost  $1.20  ;  total,  $  16.20.  20.  For  $10,  I  can  buy  2  tons ; 
for  80  ^,  I  can  buy  (80  X  4)  lb.,  or  320  lb.  23.  347  +  495  =  (347 
+  500)  —  5.  25.  One  man  can  do  -^  in  1  day ;  the  other  can 
do  :fV  in  1  day;  both  can  do  ^^  +  ^^  in  1  day,  or  A  + A-  =  A 
=  ^  in  one  day,  thus  requiring  16  days  to  do  the  whole  work. 

890.  1.   $150,  at  4%,  for  3  years. 

2.  12  cu.  ft.,  at  60  lb.  to  cu.  ft. 

3.  $12  is  3%  of  what? 

4.  250  is  what  per  cent  of  500? 
6.  12  is  what  per  cent  of  4? 

6.   20M@$30perM. 


NOTES   ON   CHAPTER   ELEVEN  105 

7.  4  bbl.,  300  lb.  each,  @  5^.         9.    84  ^  4. 

8.  18  ^4i-.  10.    75  @  $79  (or  $80). 


4.    Find  ^  and  annex  cipher. 
13.   See  Art.  791. 

893.   See  Arts.  792,  716,  717,  714. 

26.  Multiply  by  36,  by  subtracting  4  times  the  number  from 
40  times  the  number ;  multiply  by  45,  by  subtracting  5  times  the 
multiplicand  from  50  times  the  multiplicand. 

895.   See  Art.  563,  p.  55. 

897.   2.    18a:+15:r  +  18a;  +  15a:  +  (18xl5)  =  930. 

6.  Floor  space  =  (30  X  24)  sq.  ft. 
Air  space  =  (30  X  24  X  15)  cu.  ft. 

7.  (30x24)-^fJ. 

8.  Reducing   to   yards  :    [(10  X  5)  +  (8  X  5)  -f  (10  X  5)  + 
(8x5)  +  (10  X  8)]  -^  3. 

10.  Commencing  at  lower  right-hand  corner :  (15  +  3+12 
+  9  +  8  +  18  +  10  +  15  +  9  +  15)  rd. 

12.    [(22  X  12)  X  (14  X  12)  X  (9  X  12)]  -^  2150.4. 

15.  Dimensions  :   1000  yd.,  2  yd.,  3^  yd. 

16.  Dimensions  of  pipe  space  :   1000  yd.,  1^  yd.,  1|-  yd. 

900.  It  may  be  necessary  for  the  teacher  to  supplement  the 
information  given  the  pupils  in  connection  with  the  demand 
notes  in  Art.  877.  The  present  note  is  payable  at  a  fixed  time, 
and  the  place  of  payment  is  specified ;  but  it  does  not  bear 
interest.  If,  however,  it  is  not  paid  at  maturity  it  bears  interest 
from  that  time  at  the  legal  rate. 

While  savings  banks  loan  money  only  on  good  security,  gener- 
ally real  estate,  banks  of  deposit  will  advance  money  on  a  note, 
if  the  officers  feel  certain  that  it  will  be  paid  at  maturity.  When 
William  Brown  &  Sons  present  the  note  for  discount,  they  endorse 


106  MANUAL   FOR  TEACHERS 

it  by  writing  their  name  on  its  back.  This  transfers  the  owner- 
ship of  the  note  to  any  subsequent  holder,  and  also  makes  the 
endorsers  liable  for  the  amount  in  case  the  maker  fails  to  pay  it 
at  maturity.  The  discounting  bank  thus  has  two  parties  upon 
whom  to  depend  for  the  money. 

The  sum  charged  by  the  bank  for  this  service  is  the  interest 
on  the  face  of  the  note  for  the  time  it  has  to  run.  This  sum  is 
called  the  discount,  as  it  is  deducted  from  the  sum  named  in  the 
note;  and  the  difference  —  called  the  avails  or  proceeds  —  is 
given  to  the  owner  of  the  note. 

When  the  above  note  is  due,  it  is  sent  to  the  Park  National 
Bank  for  collection.  If  Thomas  Tierney,  or  some  other  person, 
does  not  pay  the  money  before  the  close  of  banking  hours,  the 
note  is  protested;  that  is,  a  notary  public  certifies  that  payment 
has  not  been  made,  and  notifies  the  endorsers,  William  Brown 
&  Sons,  of  their  liability. 

901.  In  states  in  which  days  of  grace  are  no  longer  allowed, 
the  pupils  should  not  employ  them  even  in  calculating  discount 
on  notes  made  in  places  that  still  have  days  of  grace.  Two 
answers  are  given  to  each  problem  in  discount,  one  including 
days  of  grace ;  the  other,  enclosed  in  parentheses,  in  which  days 
of  grace  are  not  employed. 

902.  These  exercises  are  nothing  more  than  examples  in 
interest,  except  that  in  some  states,  three  days  are  to  be  added  to 
the  time  mentioned. 

Deducting  the  discount  from  the  face  of  the  note  gives  the 
proceeds. 

903.  As  will  be  seen  in  Art.  906,  the  exact  number  of  days  is 
taken  for  periods  less  than  a  year. 

904.  Pupils  should  be  led  to  see  that  banks  are  entitled  to 
interest  only  for  the  number  of  days  they  have  to  wait  for  repay- 
ment. A  failure  to  understand  this,  leads  to  frequent  mistakes. 
Many  careless  scholars  find  the  difference  in  time  between  th« 


NOTES  ON  CHAPTER  ELEVEN  107 

two  dates  named  in  the  example,  —  from  Feb.  27  to  March  9,  in 
25,  — disregarding  entirely  the  time  for  which  it  is  drawn. 

Instead  of  explaining  how  to  calculate  the  discount  on  the 
notes  given  in  Art.  905,  the  teacher  should  permit  the  class  to 
attempt  to  ascertain  the  result  by  themselves.  In  case  of  a 
failure  to  obtain  the  correct  answer,  a  discussion  of  the  matter 
will  lead  to  a  proper  understanding  of  the  principles  involved. 

906.  1900  is  not  a  leap  year.  See  Arithmetic,  Art.  1303, 
Time  Measure. 

907.  Find  the  total  number  of  meters  in  the  first  twelve 
pieces,  and  ascertain  their  value  at  the  price  named.  Do  the 
same  for  the  remaining  four  pieces.  Eeduce  the  total  number  of 
meters  in  the  sixteen  pieces  to  yards,  by  multiplying  by  39.37 
and  dividing  the  product  by  36. 

90a   See  Arithmetic,  Art.  758. 

910.  In  64^  X  11-f ,  the  product  by  ^  is  found  by  multiplying 
the  product  of  4-,  already  ascertained,  by  4. 

912.  8.  A  yard  is  36  in.  If  ribbon  is  36  in.  long,  its  width 
must  be  (144  -^  36)  in.  to  contain  144  sq.  in. 

14.  Some  pupils  will  say  without  reflecting,  200%  —  not  see- 
ing that  the  profit  is  equal  to  the  cost,  100%. 

15.  1%  of  $1500. 

20.    10  =  what  per  cent  of  (40  +  10)  ? 

27.    The  remainder  =  20% ,  or  ^  =  $  2000. 

29.  The  thoughtful  teacher  must  determine  for  herself  just 
how  much  time  she  can  afford  to  waste  in  giving  the  pupils  a 
number  of  useless  facts  about  taxes,  brokerage,  commissions, 
bonds,  etc.,  etc.  The  time  allotted  to  arithmetic  should  be  spent 
chiefly  in  developing  "  power  "  in  her  scholars.  If  the  latter  can 
correctly  apply  mathematical  principles  in  ordinary  problems 
suited  to  their  present  experience,  they  will  not  find  it  difficult 


108  MANUAL    FOR   TEACHERS 

in  later  life,  after  they  understand  the  conditions,  to  solve  such 
new  problems  as  come  up. 

In  this  example,  it  will  be  sufficient  to  say  that  the  "  premium 
for  insuring  "  means  "  cost  of  insuring." 

913.  1.    See  Art.  685. 

3.    The  pupils  should  attempt  to  frame  the  definitions  asked. 

5.  (a)  fof  832  =  ^7zs. 

(h)  832  =  4  of  a:  =  ^;  etc. 
8 

6.  1^=3750. 
100 

8.  The  first  boy  gains  J^  on  a  1  ^  apple,  or  25%  ;  the  second 
gains  I  ^,  or  20%. 

9.  Sold  (20  X  20)  sq.  rd.  +  16  sq.  rd. 

15.    The  pupil  should  be  able  to  state  the  rule. 

19.    a; +  2^0;  =  1050. 

21.   ( [  (35  +  23  +  35  +  23)  x  13]  -f  [35  X  23])  ^  9. 

914.  Use  first  as  sight  problems. 

1.  2£_iQ_^. 

3  4 

2.  a:  +  Y  =  24. 

3.  a:  +  a;  +  5  =  31. 

4.  If  J  of  A's  money  =  |  of  B's,  A's  money  =  f  of  B's  X  } 

=  fB's.     Let  a;  =  B's,  ^=  A's.     a: +^=  165;  ?4^  =  165; 
5  5  5 

dividingby  ll,f=15;  a:  =  75.     B's  =  $75,  A's  =^90. 

8.    After  2  days,  there  will  be  enough  to  feed  4  horses  4 
days,  or  1  horse  16  days,  or  5  horses  3^  days. 


NOTES  ON  CHAPTER  ELEVEN  109 

915.  7.  An  ounce  avoirdupois  contains  7000  gr.  -f- 16  ;  an 
ounce  troy  contains  480  gr. 

8.  4  lb.  8  oz.  avoirdupois  =  7000  troy  grains  X  4^. 

9.  The  pure  silver  amounts  to    192.9    gr.  X.9;    divide   by 
480  to  reduce  to  ounces,  and  multiply  75^  by  the  quotient. 

75^  X  192.9  X  .9 
480 

11.  Number  of  square  feet  =  [(16  +  14  +  16  +  14)  X  8]  -f 
16  X  14. 

Note.  —  When  the  bottom  of  a  tank  is  covered  with  sheet  lead,  the  side 
strips  will  be  ^^  in.  less  than  8  ft.  high,  etc.,  but  this  small  difference  may  be 
neglected  in  these  four  problems. 

12.  Multiply  the  number  of  square  feet  by  -^^  and  divide  the 
product  by  12.     Cancel. 

21.  Assessed  value,  i  of  $24000,  or  $18000.  Taxes  =  1|% 
of  $18000. 

917.  8.   Dividend  -  2^%  of  ($50  X  65). 

918.  1.  Area  of  surface  to  be  papered :  [(16  +  14  -f  16  +  14) 
X  10]-  174  ;  divided  by  area  of  roll,  24  ft.  by  H  ft- 

4.  When  he  sells  31  gills,  the  grocer  charges  for  32  gills,  or 
1-jY  times  the  correct  quantity,  thereby  charging  for  -^  more  than 
he  gives.  In  2  hhd.  of  58  gal.  2  qt.  1  pt.  each,  there  are  117:^ 
gal.,  the  dishonest  gain  on  which  is  ^^  of  117J  gal.  worth  $.80 
X^Vof  117i. 

6.    30%  of  cost  {x)  =  21. 

919.  1.  (a)  $  48.50  +  ($  48.50  X  y^^  X  -^Vo)  =  ?  48.50  +  .51 
=  $49.01.  Arts. 

Omitting  days  of  grace,  $48.50  +  ($48.50  X  yf^  X  ^W)  = 
$  48.50 +  .48f  =  $48.98J,  say  $48.99.   Ans. 

These  examples  should  be  worked  with  days  of  grace  or  without  days  of 
grace,  but  not  in  both  ways.  See  Art.  901.  Days  of  grace  were  not  abol- 
ished in  New  York  until  January  1895. 


110  MANUAL   FOR  TEACHERS 

(b)  With  days  of  grace,  this  note  is  due  Dec.  17.  Term  Dec.  1 
to  Dec.  17  =  16  days.  Discount  on  $49.01  for  16  days  at  6% 
=  |49.01  X  T^  X  J^  -=  13^.  Proceeds  =  $49.01  -  .13  =  $48.88 
Ans. 

Omitting  days  of  grace,  the  note  is  due  Dec.  14.  Term  =  13 
days.  Discount  on  $48.99  for  13  days  at  6%  =$48.99  X  ^ 
X^=np.     Proceeds  =  $48.99 -.11  =  $48.88.  Ans. 

2.  With  days  of  grace,  the  amount  due  at  maturity  will  be 
$175  +  ($175Xt^X^)  =  $175  +  2.71  +  =  $177.714-.  The 
term  of  discount  =  93  days  —  33  days  =  60  days.  Interest  of 
$177.71  for  60  days  at  6%  will  be  $1.78  nearly.  Proceeds 
=  $177.71  -  1.78  =  $175.93.  More  accurately,  $177.7125  - 
1.7771  =$175.9354  or  $175.94-. 

Without  days  of  grace,  the  amount  due  at  maturity  will  be 
$175  4-  ($  175  X  yf^  X  ^)  =  $  177.63  - .  Interest  of  this 
amount  for  57  days  =  $1.69  —  .  Proceeds  =  $177.63- 1.69 
=  $175.94. 

920.  See  Arithmetic,  Arts.  821,  822. 

921.  9.   4)  £183  Us.Sd. 

13.  Total  number  of  days'  work  =  32-f-  53  +  41  =  126.  Value 
of  1  day's  work  =  $283.50^  126.  Share  of  first  man  =  ($283.50 
H-  126)  X  32.     Cancel. 

14.  Amount  furnished,  $  12000.  The  one  furnishing  $3000, 
or  J,  is  entitled  to  ^  of  $  1800 ;  the  second  to  ^  of  $  1800,  etc. 

15.  After  10  days,  there  are  rations  for  1200  men  for  30 
days ;  which  will  last  1  man  30  days  X  1200 ;  and  will  last  1200 
men  +  300  men,  (30  days  X  1200)  -f- 1500  =  Ans. 

16.  Train  leaving  B  goes  1^  times  as  fast  as  the  other,  so  that 
meeting  place  will  be  1^  times  as  far  from  B  as  it  is  from  A.  If 
X  represents  distance  from  A,  l^x  will  represent  distance  from 
B,  and  x-\-l\x=  120,  or  a;  =  48.  Trains  meet  48  mi.  from  A, 
or  72  mi.  from  B.  The  first  train  takes  fj  hr.  to  travel  the 
distance,  or  2  hr.  24  min. ;    second  train  requires  the  same  time, 


NOTES   ON   CHAPTER   ELEVEN  111 

Jl  hr.,  or  2  hr.  24  min.  Time  of  meeting  =  9  A.M.  -f  2  hr.  24 
min.  =  24  min.  past  11. 

Or,  let  a;  =  time  required  to  reach  meeting  point;  then  20a; 
=  distance  travelled  by  one  train,  and  30  a;  =  distance  trav- 
elled by  the  other,  and  20  a;  +  30  a;  =  whole  distance  =  120,  or 
X  ~  2|-.     Time  is  2|-  hr.,  etc. 

17.  a:  +  3a7  +  6a;  +  10a;  =  900. 

18.  Let  X  =  share  of  third  ;  a;-}-  75  =  share  of  second  ;  and 
a;  +  75  +  48  =  share  of  first. 

a;  +  a:+75  +  a;+75  +  48  =  540. 

19.  If  4  men  need  105  hr.,  one  man  would  need  420  hr.,  and 
6  men  would  need  70  hr.,  or  (70  -^  10)  da. 

20.  Let  X  =  number  of  dozen  bought,  15  a;  =  cost  in  cents  ; 
a;  —  1-^  =  number  of  dozen  sold,  16  times  (x  —  1^)  =  16 a;  —  20 
=  selling  price  ;  16  a;  -  20  =  15  ar,  or  x=-  20.  He  bought  20 
dozen,  or  240  eggs. 

.  21.  The  interest  on  $250  for  8  mo.  is  the  same  as  that  on  $  1 
for  8  mo.  X  250,  and  on  $400  for  (8  mo.  X  250)  ^  400,  or  5  mo. 

22.  Provisions  for  3000  men  would  last  1^  times  as  long  as  for 
4000  men,  or  18  wk.  X  1^  =  24  wk.  Am. 

Or,  (18  wk.  X  4000)  -^  3000. 

24.  Area  of  first  plank  in  square  feet,  20  X  1 ;  of  second, 
24  X  a;.     24  a;  =  20. 

a:  =  If  Ans.  in  feet,  or  (||  X  12)  in. 

25.  He  can  pay  |^|^  of  his  debts.  Mr.  Smith  should  receive 
$  576  X  fJIf.     Cancel. 

922.  6.    See  table.  Arithmetic,  Art.  795. 

923.  10.    23  X  11  X  a;  =  2749. 
13.   48  X  72  X  a;  =  2150.4  X  40. 

2150.4  X  40 


48x72 
22.    55  cts.  X  6  X  54  X  M- 


=  Ans.  in  inches. 


112  MANUAL   FOR  TEACHERS 

924.     1.    The  pupils  should  gradually  become  accustomed  to 
business  methods  of  obtaining  results.     In  calculating 
the  amount  to  be  paid,  a  clerk  writes  the  discount  at    'P^O'^-'*^ 
once  under  the  gross  price.     He  takes  ^  by  dividing 
by  2  and  placing  the  first  quotient  figure  one  place  to    $554.23 
the  left. 

2.  In  the  first  example,  a  discount  of  5%  is  made  for  prompt 
payment ;  the  discount  here  allowed  is  a  reduction  from  what  is 
called  the  "  list  "  price.  Catalogues  are  issued  by  some  merchants 
on  which  the  prices  named  are  not  the  ones  regularly  charged, 
but  are  much  larger  so  as  to  mislead  persons  that  do  not  know 
the  rate  of  discount  allowed.  Information  as  to  this  rate  is  com- 
municated to  customers,  and  varies  from  time  to  time  owing  to 
fluctuations  in  the  market,  the  "  list "  price  seldom  being  changed. 
The  list  price  is  sometimes  called  the  "gross"  price,  the  "net" 
price  being  the  one  actually  paid. 

A  bill  for  the  Roman  candles  would  be  made  out  as  follows : 

16  gross  Roman  Candles,    $  26. 75,    $  428.  - 

Less  60%,       256.80 

Net,    $171.20 

The  product  by  .60  is  written  under  the  "  gross"  total,  the  first 
figure  being  written  two  places  to  the  left. 

The  net  cost  can  be  directly  obtained  by  multiplying  $428 
by  .40. 

7.  Two,  three,  and  even  more  discounts  are  very  frequent  in 
business.  An  article  catalogued  at  $100  is  sold,  for  instance, 
at  $70,  and  customers  informed  that  the  discount  is  30%.  A 
later  reduction  in  price  is  accompanied  by  a  notice  that  a  fur- 
ther discount  of  10%  will  be  allowed.  This  does  not  signify 
30%  -f  10%,  or  40%,  from  the  "  list "  price  ;  it  means  that  the 
regular  price  of  $70  is  to  be  reduced  $7,  making  the  new  price 
$63.  A  third  discount  of  5%  means  5%  of  the  last  price,  $63; 
etc.,  etc.  In  writing  these  discounts,  the  per  cent  mark  is  written 
only  after  the  last. 


NOTES   ON   CHAPTER    ELEVEN  113 

11.  On  a  bill  of  $100,  40  and  10%  gives  a  "  net  "  amount  of 
$60  — $6,  or$54;  30  and  20%  gives  $70-$14,  or  $56;  the 
former  being  better  for  the  buyer  by  $2. 

12.  $100  less  33^  and  10%  ==$60.  The  discount  is  $40  on 
$100,  or  40%.     The  net  is  60%. 

13.  100-40  =  60;  20%  of  60=12;  discount  =  40%  +  12% 
=  52%.  Am. 

14.  Let  a?  =  "  list "  price.  After  first  discount  of  ^  is  de- 
ducted,  there  will  remain  — •     Deducting  10%  of  this,  or  -^  of 

o 

it,  there  will  remain  ^  of  it.     ^Sg-  of  —  =  —  =  60. 

o       lo 

15.  The  first  reduction  is  100%  -20%,  or  80%  of  the  list; 
the  second  is  100%  -  10%,  or  90%  of  the  former.  90%  of  80% 
=  Aof  80%  =  72%.  Ans. 

16.  80%  of  90%  =72%.  Ans.  The  net  price  is  the  same 
for  the  same  discounts  in  whatever  order  they  are  taken. 

925.   3.   -^  of  5000  (cents)  =  5  (cents)  ;  50a;=5;  etc. 
100 

11.   Value   at  par,   $  50  X  96  =  $  4800.      Discount  =  $4800 

-$4476  =  $324  =  a:%  of  $4800,  i.e.,  324  =  48a;;  etc. 

16.   $500Xyf^Xttf. 


3.   50%  +  [10%  of  (100%  -  50%)]  =  50%  +  5%  = 
55%.  Am. 

4.  30%  +  [30%  of  (100%  -  30%)]  =  30%  +  (30%  of  70%) 
=  30%  +  21%  =  51%.  Am. 

5.  y9^  of  gross  price  (x)  =  729  ;  ^x  =  81 ;  x  =  810. 

927.   21.    Cost  per  acre  =  $  40293 -^  396,  at  which  price  112 
acres  were  sold. 

25.  AO^X^X^.     Cancel. 

26.  Number  of  hours  =  365  -^  4^. 


114  MANUAL   FOR  TEACHERS 


92a   2.    a;Xy^Xj  =  |^-180;  9a:  =  14400;  etc. 


4.    4250  X  yf^  X  a;  =  765.  6.    2020  X  yf^  X  a:  =  606. 

100     2 


5.   a:XT^^x3  =  240.  7.    6000  x-^x^  =  900. 


929.  1.  The  pupils  should  not  be  shown  how  to  calculate 
these  areas. 

2.  If  any  difficulty  is  experienced  in  finding  the  areas  of 
these  triangles,  the  pupils  should  be  referred  to  1 ;  after  which 
they  should  be  led  to  deduce  the  rule.  Thus  the  area  of  the 
second  triangle  may  be  calculated  from  the  figure  in  1  by  adding 
i  of  (60  X  50)  to  i  of  (60  X  50)  ;  that  of  the  third  by  adding  i 
of  (60  X  60)  to  i  of  (60  X  40)  ;  and  that  of  the  fourth  by  adding 
i  of  (60  X  70)  to  i  of  (60  X  30).  Each  of  these  will  be  found 
equal  to  i  of  (60  X  100). 

3.  The  second  rectangle  is  divided  into  three  triangles,  two 
of  them  right-angled.  By  deducting  from  the  area  of  the  rec- 
tangle the  sum  of  the  areas  of  the  two  right-angled  triangles, 
they  will  obtain  the  area  of  the  remaining  triangle. 

4.  The  area  of  each  of  these  triangles  can  be  ascertained  by 
referring,  to  the  corresponding  triangle  of  3.  Let  the  scholars 
do  this  for  themselves. 

5.  The  areas  of  the  oblique-angled  triangles  constituting  the 
first  and  second  quadrilaterals,  can  be  calculated  by  the  pupils 
that  have  benefited  by  the  work  in  4.  If  they  see  that  the  area 
of  each  triangle  of  a  parallelogram  is  equal  to  -J-  (base  X  altitude), 
the  area  of  the  latter  is  equal  to  base  X  altitude. 

For  definitions  of  quadrilaterals  see  Art.  1265. 

6.  The  area  of  the  first  is  equal  to  the  area  of  the  rectangle, 
(50  X  60),  plus  the  area  of  the  triangle,  i  of  (50  X  60)  ;  or  4500 
sq.  ft. 

The  second  is  made  up  of  a  rectangle  and  of  two  triangles ; 
its  area  is  also  4500  sq.  ft. 


NOTES  ON  CHAPTER  ELEVEN  115 

The  pupils  should  be  led  to  see  that  if,  in  the  fourth,  the  upper 
left  triangle  were  cut  off  and  placed  below,  and  if  the  lower 
right  triangle  were  cut  off  and  placed  above,  as  indicated  by  the 
dotted  lines,  the  resultant  figure  would  be  a  rectangle  60  X  75. 

Cutting  off  both  triangles  in  the  third,  and  placing  them  above, 
will  make  a  60  X  75  rectangle. 

The  area  of  each  trapezoid  is  equal  to  [^  of  (50  -f  100)]  X  60. 

7.  The  area  of  each  of  these  quadrilaterals  equals  ^  of  (30  X  100) 
+  i  of  (40  X  100),  or  [-1-  of  (30  +  40)]  X  100. 

The  first  three  quadrilaterals  are  trapeziums.  The  last  is  a 
trapezoid.     Which  are  the  parallel  sides  ? 

8.  A  strip  of  paper  of  any  uniform  width  may  be  used. 
Carefully  cut  a  rectangle  by  making  square  corners  with  a  card. 
Using  the  base  of  the  rectangle  as  a  measure,  place  two  dots  on 
the  lower  edge  of  the  strip  to  mark  the  extremities  of  the  base  of 
a  parallelogram  equal  in  length  to  the  base  of  the  rectangle,  and 
above  these,  at  any  convenient  distance  to  the  right  or  to  the  left, 
two  others  to  mark  the  extremities  of  the  opposite  side  of  the 
parallelogram.  Draw  lines  forming  the  right  and  left  sides,  and 
cut  along  these  lines.  That  the  parallelogram  is  equal  in  area 
to  the  rectangle,  may  be  shown  by  carefully  drawing  a  perpen- 
dicular at  one  corner;  cutting  off  the  triangle  thus  made,  and 
placing  it,  in  a  reversed  position,  on' the  opposite  side  of  the 
parallelogram. 

9.  See  6,  third  and  fourth  trapezoid. 

930.  2.  Four  faces  will  measure  6  ft.  by  4|-  ft.  each,  and 
two  will  measure  4|-  ft.  by  4|-  ft.  each. 

4.  Dimensions  of  floor,  57  ft.  by  18  ft.,  or  19  yd.  by  6  yd. 

5.  Volume  in  cubic  feet,  4  X  4  X  12.  Multiply  by  1000  to  get 
the  weight  in  ounces  of  an  equal  volume  of  water.  Multiply  by 
2.8  to  get  weight  of  marble  in  ounces.  Divide  by  16  X  2000 
to  reduce  to  tons. 

4  X  4  X  12  X  1000  X  2.8 
16  X  2000 


116  MANUAL   FOR  TEACHERS 

9.  Outer  dimensions,  14  X  14  X  14,  or  2744  cu.  in.  See  if  the 
same  number  of  cubic  inches  of  wood  is  obtained  by  calculating 
the  volume  of  the  wood  in  6  —  2  pieces,  12  X  12,  1  in.  thick ;  2 
pieces,  12  X  14,  1  in.  thick ;  2  pieces,  14  X  14,  1  in.  thick. 

10.  A  cube  of  water  2  ft.  long  contains  (2x2x2)  cu.  ft.,  or 
8  cu.  ft.  At  1000  oz.  to  a  cubic  foot,  it  weighs  J^^  lb.  X  8,  or 
500  lb.  The  cube  of  iron  weighs  8  times  as  much  as  an  equal 
volume  of  water. 

A  cube  of  iron  1  ft.  long  also  weighs  8  times  as  much  as  a 
corresponding  cube  of  water,  or  8  times  — J-J^  lb.  =  500  lb., 
or  \  ton. 

A  3  ft.  cube  of  iron  contains  27  cu.  ft.,  weighing  8  times  as 
much  as  a  corresponding  cube  of  water,  or  216  times  1000  oz. 
=  6J  tons. 


XV 

NOTES   ON   CHAPTER   TWELVE 

931.  While  problems  requiring  the  pupil  to  find  the  principal, 
the  rate,  or  the  time  have  very  little  "  practical "  value,  they  can 
be  so  readily  taught  by  the  algebraic  method  that  the  time  spent 
upon  them  need  not  be  very  great.  A  pupil  that  is  able  to 
calculate  one  of  a  series  of  related  items  is  benefited  by  being 
required  to  calculate  the  others,  if  he  is  not  compelled  to  resort 
to  a  series  of  ill-understood  rules  in  order  to  obtain  the  results. 

Although  there  is  no  real  difference  between  the  algebraic 
method  and  the  arithmetical  one,  a  great  number  of  scholars  fail 
to  obtain  a  thorough  understanding  of  the  latter.  They  can 
work  a  number  of  examples,  following  a  model  solution  at  the 
head  of  the  lesson ;  but  they  fail  to  grasp  the  underlying  prin- 
ciples. By  the  algebraic  method,  x  is  used  to  represent  the 
number  of  years  or  dollars,  or  the  rate,  instead  of  the  1  year,  %  1, 
or  1%,  of  the  other ;  but  this  method  seems  to  require  the  formu- 
lation of  a  number  of  rules,  as  against  practically  none  in  the 
case  of  the  other. 

After  pupils  have  learned  to  work  examples  by  the  algebraic 
method,  they  can  be  encouraged  to  discontinue  the  use  of  the  x ; 
but  they  should  not  be  taught  both  methods  at  one  time. 

933.  Represent  the  required  item  by  x.  Simplify  the  first 
member  before  proceeding  to  solve  the  equation. 

1.  2000  X -^  X  3  =  300. 

100 

2.  1800  X  yJt^  X  a:  =  144. 

117 


118  MANUAL   FOR  TEACHERS 

3.  xX-^jfX^  =  2.88. 

4.  38_|.38xa;x2^^Q28. 

Or,  38  X  -^  X  2  =  40.28  -  38  =  2.28. 

5.  U0x-^xiU  =  x  =  An8. 

a:x4x5^3g^Q 
^  100x2 

4gO_i-i52_>ll2i£  =  484.15. 
200 

15.    Principal  =  $97.57 -$7.57  =  $90. 
90  X  y^  X  a;  =  7.57. 

21.  Let  a;  =  principal. 

,  a:  X  4  X  846  _  195  92^ 
^  100x360 
The  interest  is  then  found  by  subtracting  from  $  196.92,  the 
value  obtained  for  x. 

22.  First  find  the  principal  (x). 
25.    See  15. 

934.  The  recommendation  so  frequently  made,  that  all  writ- 
ten work  be  preceded  by  oral  exercises  of  the  same  character,, 
should  not  be  followed  without  some  modifications.  Oral  work 
is  necessarily  accompanied  by  a  number  of  devices  that  tend  to 
simplify  the  task  of  handling  numbers  that  are  not  seen ; 
written  work  should  follow  general  rules  in  order  to  be  learned 
by  a  majority  of  the  pupils,  although  later  they  may  adopt  some 
of  the  short-cuts  of  their  oral  exercises.  Even  the  oral  addition 
of  two  numbers  of  two  figures  each,  is  done  in  a  manner  different 
from  the  ordinary  slate  method,  the  operation  in  the  former  case 
being  commenced  generally  with  the  tens*  figures,  and  in  the 


NOTES  ON    CHAPTER   TWELVE  119 

latter  case  with  the  units'  figures.  The  reduction  to  a  common 
denominator  recommended  in  oral  division  of  fractions,  is  seldom 
employed  in  slate  work. 

The  average  scholar  is  able  to  handle  "  mental  "  problems  con- 
taining small  numbers  in  a  way  that  he  cannot  always  explain, 
although  he  may  endeavor  to  stultify  himself  by  repeating  a 
prescribed  form  of  analysis.  It  is  next  to  impossible,  with  the 
average  teaching,  to  get  the  same  pupil  to  work  some  varieties  of 
"  written  "  problems  containing  the  same  conditions. 

In  order  to  furnish  a  general  method  of  treating  some  classes 
of  examples,  it  has  been  thought  best  to  commence  with  written 
work,  leaving  the  mental  exercises  with  their  various  devices 
until  the  former  task  is  accomplished. 

The  accomparving  exercises  are  so  simple  as  not  to  need  ex- 
planation by  the  teacher  ;  but  sufficient  time  should  be  given  the 
pupil  to  work  them  out  in  his  own  way.  They  differ  in  this 
respect  from  the  oral  examples  of  a  single  operation  containing 
larger  figures,  but  which  do  not  require  any  effort  on  the  part  of 
the  scholar  to  determine  which  process  is  required. 

1.  Yearly  interest  is  $6;  a  year  and  a  half  will  be  needed 
to  make  the  interest  $  9. 

2.  The  yearly  interest  is  $  8,  making  the  rate  4%. 

3.  Yearly  interest  is  $4,  requiring  a  principal  of  f  100,  at 
the  given  rate. 

5.  The  pupils  may  remember  (Art.  878,  No.  15)  that  6%  for 
a  year  is  1  %  for  72  da. 

6.  4%  per  year  is  1%  for  90  da. 

11.  2  mo.  12  da.  =  72  da.     See  5. 

12.  1%  for  80  da.  is  (360  -^  80)  %  for  a  year. 

17.  2%  for  6  mo. 

18.  $  3.60  per  year  is  1  cent  per  day. 
20.    See  18. 


120  MANUAL   FOR  TEACHERS 

935.  First  payment  =  -,  leaving  —  remainder ;  second  pay- 

ment  =  -J-  of  -—  =  -,  leaving  -  remainder  ;  third  payment  =  f  of 

^  =  ^  ;  last  payment,  =  $  2000.    The  total  cost  of  the  house,  x  = 
3     5 

the  sum  of  the  payments,  q  +  o  +  H  "^  2000. 

o        o        O 

936.  The  books  contain  many  methods  of  calculating  interest, 
but  it  is  questionable  whether  it  is  not  time  wasted  in  giving  so 
much  attention  to  this  topic.  The  average  person  is  required 
to  do  comparatively  little  work  in  this  line ;  while  those  called 
upon  to  compute  interest  often,  learn  short  methods  of  their  own 
or  use  interest  tables. 

If  a  second  method  is  to  be  taught  at  all,  the  one  by  aliquot 
parts  is  the  most  useful,  as  modifications  of  this  method  may  be 
applied  to  other  operations. 
6.    See  Arithmetic,  Art.  384. 

937.  A  modification  of  the  so-called  "  60-day  method." 
16.    See  Art.  901  as  to  days  of  grace. 

938.  21.  10%  gives  2  years'  interest;  then  1  yr.  (J  of  the 
foregoing)  ;  6  mo. ;  1  mo.;  18  da.  (^^  of  6  mo.). 

942.  46.   Term,  57  da.  (54  da.). 

47.  Term,  92  da.  (89  da.).  49.    Term,  34  da.  (31  da.). 

48.  Term,  16  da.  (13  da.).  50.    Term,  187  da.  (184  da.). 

943.  9.   See  Table,  Arithmetic,  Art.  1303. 

944.  11.   The  net  price  of  goods  catalogued  at  x  dollars,  and 

sold  at  a  discount  of  20  and  10%,  will  be  (^ -^>  or  ^\ 
_/.      ^80a;\_80a:      8a:  _  72  a;  ^  ^ 

y^^  100  J    100    100    100  ■ 


NOTES   ON   CHAPTER  TW! 


13.  If  tlie  selling  price  of  the  above  is  $360,^=360; 
12  x^  36000 ;  x  =  500.     Catalogue  price  =  $  500.  Am. 

14.  750  -  (-^  of  750)  =  500 ; 

500  -  C"^  of  500^  =  500  -  5  a;  =  net  price. 

500 -5a;  =  450. 
Transposing,  —  5  a:  ==  —  50. 

Changing  signs  of  both  terms,  5  a;  =  50, 

a;=10. 

945.   7.    Let  x  —  selling  price  of  muslin. 

(84  X  40)  +  105  a:  =--  (84  X  55)  +  (105  X  20). 

Another  way  :  He  loses  15^  per  yard  on  84  yd.,  which  is  a 
loss  of  15^  X  84.  This  he  must  make  up  on  105  yd.,  which  is 
(15^  X  84)  -T- 105  on  each  yard,  or  12^.  Selling  price  of  muslin, 
20)2^+12^,  or  32^.  Ans. 

8.  •J  of  them  brought  $120;  \  of  remainder,  or  J  of  them, 
brought  $96;  -J-  of  remainder,  or  J  of  them,  brought  $40;  re- 
mainder, or  i  of  them,  brought  $30.  Total  amount  received, 
$286. 

9.  Proceeds  of  gas  stock,  $25  X  165  =  $4125.  Cost  of  lots, 
$4125  -  $27  =  $4098.  Number  of  square  feet  in  lots,  (32  X  115) 
+  (30  X  105)  =  3680  +  3150  =  6830.  Value  per  square  foot, 
$4098 -^6830  =  $0.60.  Ans, 

10.  Two  walls,  each  16  X  14,  and  two  others,  each  12  X  14, 
contain  (32  +  24)  X  14,  or  (56  X  14)  sq.  ft.  =  784  sq.  ft.  The 
ceiling  contains  (16  X  12)  sq.  ft.  =  192  sq.  ft.  Adding  this  to 
the  walls,  makes  a  total  of  976  sq.  ft. 

The  deductions  are  (8  X  4)  sq.  ft.  X  2,  and  (7  X  3)  sq.  ft.  X  3, 
or  64  sq.  ft.  +  63  sq.  ft.  =  127  sq.  ft.  Number  of  square  feet  to 
be  plastered  =  976  -  127  =  849.  Cost  at  ^^  per  square  foot 
=  2^X849  =  $16.98.  Ans. 


122  MANUAL   FOR   TEACHERS 

946.   1.    A  can  do  \  of  the  work  in  1  hr.,  and  B  can  do  |  of 

it  in  1  hr. ;  together  they  can  do  in  1  hr.  (|  + 1)  of  the  work, 
or  ^  of  it ;  and  to  do  the  whole  work  it  will  take  as  many  hours 
as  -^  is  contained  times  in  1. 

l^«  =  lxff  =  ff-2H.     Am.2^houT8. 

2.  Commission  of  2^%  ^  :^  of  amount  collected  =  $1.60. 
Amount  collected  =  $  1.60  X  40  =  $64.  Amount  remitted  =  $64 
-$1.60  =  $62.40.  Ans. 

3.  i%  of  (f  of  $  12000)  =  i%  of  $9000  =  J  of  $90  = 
$22.50.  Ans. 

Note.  —  It  may  be  advisable  to  explain  to  the  pupils  that  property  is 
seldom  insured  for  its  full  value,  because  it  is  not  Hkely  that  a  fire  will 
completely  destroy  a  building,  and  insurance  companies  reimburse  the  per- 
son insured,  only  to  the  extent  of  his  loss. 

4.  32xa;  =  6x4;  32a;  =  24;  a;  =  |f  =  f.  Ans.  f  yd.  or 
27  in. 

5.  5%  for  360  days  =  1%  for  72  days  =  2%  for  144  days. 
2%  of  $87  =  -4n5. 

6.  2%  of  $176. 

7.  Let  a;  =  commission  ;  40  a:  =  amount  invested;  a:-f40a: 
=  41  a;  =  8200 ;  x  =  200.     Ans.^  200. 

8.  $500  is -J  of  cost,  $4000. 

9.  Let  a;  =  loss,  or  20%  of  cost;  5a;  =  cost;  5a;  — a;  =  4a; 
=  selling  price. 

X,  the  loss,  is  i  of  selling  price,  4  x. 

10.  Let  a;  =  gain,  which  is  20%,  or  ^,  of  the  cost  of  the 
goods  ;  5  a;  =  cost ;  5  a;  -|-  a;,  or  6  a;,  =  selling  price. 

X,  the  gain,  is  ^  of  selling  price,  6a;. 

Note.  —  The  amount  of  money  given  in  these  two  examples,  $1200,  does 
not  affect  either  result.     It  may  be  used  or  not,  as  the  pupil  prefers. 

11.  3  men  earn  $72-5-8  in  one  day,  or  $3  per  day  each.  5 
men  earn  $15  a  day,  or  $165  in  11  days. 


NOTES   ON   CHAPTER   TWELVE  123 

12.  3  quarters  of  the  cost,  or  ^,  =  225.     Cost  =  |300.     By 

selling  for  $325,  there  is  a  gain  of  $25,  or  -^  of  the  cost.     -^^ 
=  8J%.  Arts. 

13.  2|  yd.,  orf  yd.  cost  40^;  1  yd.  costs  40 ^-^f,  or  f  of 
40^  =  Ib^.     4  yd.  1  ft.,  or  4^  yd.,  cost  15^  x  4^. 

947.    The  following  is  the  solution  without  days  of  grace  : 
Let  X  =  face  of  the  note. 


Then,  ^xAx^  =  ^  =  dkcount; 

X =  proceeds  =  1000. 

200      ^ 

200  a; -a;  =  200000, 

199  a;  =  200000, 

Face  of  note  =  $  1005.03.  Arts. 

Proof.     Face  of  note,  $  1005.03  - 

Deduct  30  days'  discount,     i%,  5.03  — 

Proceeds,  $1000.00 

949.  1.    When  days  of  grace  are  omitted,  the  term  of  dis- 
count is  90  da. 

10.    Find  the  term,  and  add  the  number  of  days  to  March  15. 

950.  2.    (a)  1  trillion,  500  billions,  etc. 

5.  The  first  quarter  of  1888  contained  (31  +  29  +  31)  da.,  or 
91  da.  The  man  was  employed  60  da.,  and  unemployed  31  da. 
His  $3  additional  paid  the  expenses  of  the  working  days. 
Deducting  $  2  X  31,  or  $  62,  for  the  expenses  of  the  other  days, 
his  net  income  =  $  350  -  $  62  =  $  288.  Ans. 


124  MANUAL   FOR  TEACHERS 

6.    Apr.  16,  79  to  Mch.  19,  '86,  83^  mo.,  @  $8,  $664.80 

Mch.  19,  '86  to  Mch.  4,  '87,  llj       "      "    12,  138.00 

Apr.  16,  '79  to  Sept.  1,  '80,  1^      "      "     2,  33.00 

Apr.  16,  '79  to  Nov.  22,  '82,  43^      "      "     2,  86.40 

Ans.  $922.20 

10.    Last  quarter's  salary  =  $287  =  82%  of  previous  quarter's 

salary  =  ^  =  2S7;  x  =  350. 
100 

Salary  of  three  quarters  @  $  350  =  $  1050 ;  add  last  quarter's, 
$287.     Total  for  year,  $1337.  Ans. 

9SL  See  Art.  784. 

953.  To  find  19  times  91,  subtract  91  from  20  times  91,  or 
1820-91. 

82  X  19  =  (82  X  20)  -  (82  x  1). 
51  X  29  =  (51  x  30)  -  51. 
27  X  99  =  2700  -  27. 

954.  See  Art.  706.     675  -^  37|  =  6f  -4-  f  =  54  -^-  3. 

955.  136  X  J  =  136  -  (^  of  136)  =  136  -  17. 

290    X      1^  =  290-29. 

64^^   5     =12^=12^. 

22    Xl9|   =  (22  X  20)  -  (22  X  ^). 

45    X    9||  =  (45xl0)-(^of45). 
160   ^    1^   =320h-3. 

18^^   1|   =   94-^9  =  10f 

956.  3.  At  50^  each,  the  cost  would  be  $8;  at  1^  apiece 
less,  the  cost  is  $8.00  -  $.16  =  Ans. 

4.   (I  of  100  lb.)  X  27  =  100  lb.  X  (f  of  27)  =  100  lb.  x  V 
=  100  lb.  X  20^. 

Note.  —  The  100  should  not  be  used  until  the  end;  even  then,  20^  is 
changed  to  2025  without  thinking  of  multiplication,  J  being  considered  25, 
and  annexed  to  20.    See  Art.  649. 


NOTES   ON    CHAPTER   TWELVE  125 

6.  900  ^  75  =  (900  -f- 100)  ^  (75  ^  100)  =  9  -^-  j. 

7.  See  Art.  955. 

9.  (10i-x4)+iofl0i  =  42  +  5i  =  47i. 

12.  i  of  (33  X  42)  =  33  X  21. 

15.  $16i^$li  =  65-f-5. 

20.  16ix2i-  =  (16ix2)  +  (iofl6i). 

957.  1.   -i^5^of(27/x56x37ixf|). 

2.  [($4,875  X  17350)  -^  196]  +  [$4.9375  x  122.75]  +  [$ .0825 
X  2240  x  2i]. 

3-    :r-"^a:  =  49739.55f. 
12.    Duty  -  [^  of  (55  ^  x  45  x  38)]  +  [20)^  x  (45  X  38  X  ff  )]• 

958.  In  multiplying  by  427,  the  first  figure  of  the  product  by 
42  (7  X  6)  is  placed  under  the  2;  in  multiplying  by  832,  the  first 
figure  of  the  product  by  8  is  placed  under  the  8,  and  the  first 
figure  of  the  product  by  32  (8  X  4)  is  placed  under  the  2. 

959.  These  exercises  contain  some  examples  worked  by  short 
methods  explained  in  previous  chapters.  See  Arts.  650,  714, 
791,  792,  and  891. 

960.  See  Arithmetic,  Art.  384. 

964.     2.   It  won  17  games  out  of  30. 

3.  1600%. 

4.  "I  is  what  per  cent  of  -i^?  ^  is  what  per  cent  of  ■§-  ?  y^  is 
what  per  cent  of  ^^  ?  5  is  what  per  cent  of  6  ?  f  =  SS^% . 

6.  The  deduction  of  the  first  discount  leaves  80%  of  the  list 
price;  the  deduction  from  this  of  10%  of  itself  leaves  90%  of 
80%,  or  72%. 

7.  One  fills  -J-  of  tank  in  1  hr.,  the  other  fills  ^  in  1  hr. ;  both 
together  fill  ^  +  ^  in  1  hr.,  or  -^  +  A'  ^^  A-i  ^^  fill  fl  of  tank, 
it  will  take  24  hr.  h-  7  =  34  hr. 


126  MANUAL   FOR   TEACHERS 

8.  6%  for  60  da.  =  80^;  for  12  da.,  ^  of  80^  or  16^;  for 
72  da.,  80^  +  16)^^-96^.  Or,  $4.80  for  year,  and  ^  of  $4.80 
for  72  da. 

9.  16%=  A;  420  =  1^;  etc. 

965.  5.  Selling  price  =  |  of  $1.50  =  $1.12| ;  gain  =  22J^ 
=  J  of  90^. 

6.  Selling  price  =  $9.60,  a  reduction  of  $2.40  from  marked 
price,  or  i  of  $12,  or  20%. 

7.  The  rug  is  sold  for  $24.  If  this  is  f  of  marked  price,  the 
latter  is  $30. 

8.  See  7. 

Note.  —  It  is  not  to  be  expected  that  all  the  pupils'  work  will  be  short- 
ened to  this  extent,  but  the  majority  of  the  class  should  be  able  to  give 
answers  at  sight  to  these  four  examples. 

9.  Find  y^  ^^  ^^^  2s.  6c?.  by  compound  division  ;  do  not 
reduce  to  pence. 

21  X 

966.  4.    Let  X  =  profits  first  year ;  then  — — -  =  profits  second 

91  T  2^ 

year;  :r  4-^  =  6970. 

5.  I  wish  to  gain  15%  of  $.96,  or  $.14|-,  which  makes  my 
selling  price  $.96  +  $.14|  =  $1.10f 

Let  X  =  marked  price. 

._15£  =  1.104;  etc. 

Or,  writing  all  the  foregoing  in  one  equation : 

100    Vioo        J 

85  a;  =  .96x115, 
^  .96  X  115 
^  =  —85— 


NOTES   ON   CHAPTER   TWELVE  127 

7.  The  average  pupil  will  be  able  to  obtain  the  meanings  of 
these  terms  by  inquiries  of  his  parents,  friends,  etc. ;  and  he  will 
remember  much  longer  what  he  learns  in  this  way,  than  if  he 
finds  the  answer  in  the  text-book.  As  the  penalties  for  taking 
usurious  interest  vary  in  the  different  states,  the  teacher  should 
ascertain  the  law  of  her  own  state  in  this  matter.  See  Art.  1306. 
A  tax  bill  or  a  policy  of  insurance  brought  in  by  a  pupil  and 
described,  will  add  to  the  interest.  The  teacher  should  not  spend 
too  much  time  upon  details  that  have  no  relevancy  in  her  sec- 
tion of  the  country.  Poll  taxes,  for  instance,  should  not  be 
dwelt  upon  in  cities  in  which  they  are  not  collected  ;  etc. 

9.  The  use  of  the  hogshead  as  a  measure  of  63  gal.  is  fast 
becoming  obsolete.  The  term  "  barrel,"  to  indicate  31^  gal.,  is 
occasionally  used  in  giving  the  capacity  of  large  tanks,  etc.  The 
U.  S.  authorities  require  prices  to  be  stated  in  the  currency  of  the 
country  from  which  the  articles  are  exported ;  but  as  this  would 
make  the  problem  more  complicated,  the  text-books  generally 
give  prices  in  U.S.  money.  No  allowance  is  now  made  for 
leakage,  the  quantity  actually  imported  being  ascertained  by 
measuring. 

10.  A  port  of  entry  is  a  place  in  which  there  is  a  custom 
house,  established  by  the  government. 

967.  4.  Any  principal  — 1 150,  $575,  or  |343.75  — will 
double  itself  at  5%  in  (100  ^  5)  yr. 

12.  The  pupils  will  need  to  obtain  a  correct  idea  of  the  mean- 
ing of  the  word  "  premium  "  in  this  connection,  as  they  will  find 
it  used  differently  when  they  come  to  the  study  of  Bonds  and 
Stocks.  The  premium  is  the  amount  paid  to  the  company  assum- 
ing the  risk. 

968.  11.  $3500  is  raised  on  property  worth  $1750000;  the 
rate  is  $3500  ^  $1750000  =  2  mills  on  $1.  The  man's  property 
tax  =  2  mills  X  24000  =  $48;  adding  to  this  1  poll  tax,  at  $2, 
gives  his  total  tax  of  $  50.  Ans. 


128  MANUAL   FOR  TEACHERS 

969.  2.  Multiply  the  denominators  of  the  first  and  the 
third  ;  divide  the  numerators  of  the  second  and  the  fourth. 

5.  While  pupils  in  lower  grades  may  be  permitted  to 
reduce  both  amounts  to  pence,  it  is  now  time  to  use  a  shorter 
method.  The  sums  given  may  be  changed  to  38Js.  and  5Js.,  or 
J-Ps.  and  -^s. 

970.  1.    See  diagram,  Arithmetic,  Art.  897,  Problem  2. 

18  X  +  15  a;  +  18  :r  + 15  a:  -f  (18  X  15)  =  63  X  9, 
66  a; +  270  =  567, 

66  a; -567 -270  =  297, 

x  =  ^.  Am.  ^  ft. 

The  arithmetical  solution  is  apparent  from  the  foregoing.  The 
sides  and  the  bottom  contain  63  sq.  yd.,  or  567  sq.  ft.  The  bot- 
tom contains  (18  X  15)  sq.  ft.,  or  270  sq.  ft.  The  sides  contain 
567  sq.  ft.  -270  sq.  ft.  =  297  sq.  ft.,  which  is  the  area  of  four 
rectangles,  whose  bases  measure  18  ft.,  15  ft.,  18  ft.,  and  15  ft., 
respectively,  the  total  being  66  ft.  297-^66  gives  4^  as  the 
number  of  feet  in  depth. 

2.  There  are  4840  sq.  yd.  in  an  acre.  (4840  X  3)  -?-  (|  of  242). 
Cancel. 

3.  The  area  of  a  trapezoid  is  found  by  multiplying  one-half 
the  sum  of  the  parallel  sides  by  the  perpendicular  distance 
between  them.  See  diagrams,  Arithmetic,  Art.  929,  Problems  6 
and  9 ;  and  Art.  1265,  Figs.  9  and  10. 

4.  The  area  =  (^^^^)  X  60  =  30  a;  +  3000.  Am. 

30  a;  +  3000  =  5400 ;  a;  =  80.     Am.  80  yd. 

5.  (m^y,=mx.Am. 

100  a:  =  4000 ;  a:  =  40,     Am,  40  yd. 


NOTES   ON   CHAPTER   TWELVE  129 

6.   /'£±|±i5Ax60  =  (a;  +  20)x60  =  60:r+1200.  Am. 

60 a; +1200  =  6000;  a;  =  80;  a;  +  40  =  120. 
Am.  80  yd.  and  120  yd. 

9.  The  number  of  square  feet  in  the  walls  of  a  room  16^  ft. 
long,  14f  ft.  wide,  and  13J-  ft.  high,  may  be  obtained  by  adding 
the  bases  of  the  four  sides,  — 16^  +  14f  -f  16^  +  14|  =  62|,  —  and 
multiplying  this  by  their  common  height,  13^.  Dividing  by  9 
gives  the  number  of  square  yards.  The  operation  of  finding  the 
cost  may  be  indicated  as  follows : 

10^X125X40^25000^^^^^^^     ^n..  $9.26. 
9x2x3  27  -^        ^  ^ 

10.  (20  X  17-J-)  -^-  2^  gives  the  number  of  feet  of  carpet.  Di- 
viding this  result  by  3  gives  the  number  of  yards. 

11.  41i  lb.  X  (15f  X  H  X  i).     Cancel. 

12.  The  "development"  will  be  a  modification  of  the  one 
given  in  problem  20,  Arithmetic,  Art.  818.  In  drawing  the 
development,  the  pupil  should  be  required  to  approximate  the 
proper  proportions,  and  to  place  the  faces  in  the  proper  order. 
It  is  not  necessary  to  have  the  top  and  bottom  faces  in  the  posi- 
tions shown  in  Art.  818. 

The  surface  of  the  four  vertical  faces  should  be  obtained  in 
one  operation,  as  in  9  ;  also  the  surface  of  the  two  horizontal  faces : 

[2  X  (3f  +  2i)  X  li] -f- [2  X  (3i  X  2^)]. 

13.  (135^-^12J)ft. 

14.  [(128  X  152  X  105)  -^  2150.4]  bu. 

15.  [(77x45x54)  ^231]  gal. 

16.  40  acres  =  (160  X  40)  sq.  rd.  =  6400  sq.  rd.  The  dimen- 
sions are  80  rd.  by  80  rd.,  making  320  rd.  of  fence  necessary. 
There  will  be  640  posts,  at  15^  each;  and  5  times  640  rails,  or 
3200  rails,  at  10^  each. 


130  MANUAL   FOR   TEACHERS 

17.  There  will  be  16  fields,  4  rows  of  4  fields  each.  Five 
parallel  fences,  each  a  mile  long,  and  five  other  parallel  fences  of 
the  same  length,  and  perpendicular  to  the  first,  will  be  required. 

19.  1728  cu.  in.  of  water  weigh  1000  oz. ;  1  cu.  in.  weighs 
(1000  ^  16)  lb.  -^  1728 ;  231  cu.  in.  weigh  [(1000  X  231)  ^ 
(16  X  1728)]  lb. 

21.  Number  of  square  feet  =  (320  +  210  +  320  +  210)  X  6. 
Divide  by  9  to  obtain  square  yards. 

22.  Area  of  outer  rectangle  in  square  feet  =  332  X  222 ;  of 
inner  rectangle  —  (320  X  210)  sq.  ft.  Divide  the  difference  by  9 
to  obtain  square  yards. 

23.  Area  of  outside  plot  =  (320  X  210)  sq.  ft. ;  of  inner  plot 
=  (308  X  198)  sq.  ft. 

972.  1.  Each  yard  measured  with  the  short  yardstick  con- 
tains If  yd. ;  the  true  length  =  25  yd.  X  ff . 

Or,  each  so-called  yard  is  ^^  yd.  short,  and  25  yd.  are  ff  yd. 
short ;  and  the  piece  contains  25  yd.  ~  -||-  yd.  =  24^  yd.  Ans. 
4.  32  boys  =  20  men.  If  15  men  do  the  work  in  12  da.,  20 
men  do  it  in  12  da.  X  \^. 

20.  Change  6  lb.  14  oz.  and  23  lb.  12  oz.  to  ounces,  or  to 
pounds  and  fractions. 

976.  3.  A  and  B  can  mow  4-  of  the  field  in  1  da.,  all  three 
can  mow  -J-  of  the  field  in  1  da.  C  mows  (^  —  ^)  of  the  field  in 
1  da.,  or  -^  of  it ;  in  5  da.  he  does  -^  X  5,  or  -J  of  the  work,  for 
which  he  should  receive  ^  of  $25. 

4.  5bu.(g  80)2^ -400^. 

5  bu.  (§60)2^-300^. 
a;bu.  @  30^  =  30 a;^. 
(x+  10)  bu.  @  50^  =  (30a;  +  700)^, 
50  a: -1-500  =  30  a: +  700, 
20  a;  =  200, 

X  =  10.  Ans.  10  bu.  of  oats. 


NOTES   ON   CHAPTER    TWELVE  131 

7.  Total  cost  -=  (65 ^  X  128)  +  80  ^  ;  quantity  sold  =  128  gal. 
-  16  gal.  Selling  price  per  gallon  =  f  of  [(65^  X  128)  +  80^] 
-^(128-16). 

It  is  advisable  to  accustom  children  to  understand  that  a  gam 
of  ^  of  cost  makes  the  selling  price  -|  of  cost. 

10.  On  sofas  sold  for  $1125  there  was  a  loss  of -|-,  making  the 
selling  price  -|  of  cost.  On  the  remaining  sofas,  the  selling  price, 
$1125,  represents  f  of  cost. 

4  fifths  of  cost  =  $1125,  selling  price  of  25  sofas. 
1  fifth  of  cost  =  $2811,  loss  on  first  lot. 
6  fifths  of  cost  =  $1125,  selling  price  of  remaining  sofas. 
1  fifth  of  cost  =  $  187^  =  gain  on  second  lot. 
Loss  on  the  transaction,  $281.25  -  $  187.50. 

The  following  is  an  algebraic  solution  without  fractions : 
Let  X  =  loss  on  first  lot ; 

then  5  a;  =  cost  of  first  lot. 

^x-~x  =  4:X  —  selling  price  =  1125. 
X  =  281^  =  loss  in  dollars. 
Let  X  ■=  gain  on  second  lot ; 

5  a:  =  cost  of  second  lot. 
hx-\- x='^x  =  selling  price  —  1125. 
X  =  187-|-  —  gain  in  dollars, 
etc. 

11.  Let  X  —  cost  per  egg. 

18  a;  =  cost  of  18  eggs. 

— ^  =  selling  price  per  egg. 

18  X  1 X  '  T      ' 

Gain  per  egg  = x  =  —-;  that  is,  on  x  cents  I  gain  -^ 

of  X  cents.     A  gain  of  -^  of  cost  =  -^^^i  of  cost  =  63^^%.  Ans. 


x-^  =  2. 


132  MANUAL   FOR  TEACHERS 

12.    Let  X  =  marked  price. 

100 

18.    Let  X  ==  cost  of  the  horse. 

"^  "^  20  ""  ^^  ^^^  ^^'^®  ""  "20" 
He  sold  it  at  ^  of  asking  price  ;  i.e.,  ■—  of  — ^,  or  — -^. 

399^^275. 
400 

20.  Length  of  field  in  rods  =  (1600  +  146)  ^  18. 

21.  Let  10 ar  represent  cost,  then  loss  =  a:;  and  selling  price 
-^  9  a;  =  $  117  ;  a:  =  13  ;  cost  =  $  130.  A  gain  of  10%  would  be 
$13,  making  the  price  at  which  he  should  have  sold  it  to  gain 
10% -$130 +  $13  =  $143.  Ans. 

22.  Wife  receives  — -;  son,  «  ^^o'  °^  ~q"'  ^^-ugli^^r,  $5000. 

^+^+5000  =  a:. 

977.  The  ten  problems  of  this  section  will  call  for  no  special 
treatment.  An  occasional  problem  of  this  kind  has  already  been 
given,  although,  perhaps,  with  smaller  numbers.  After  the 
pupils  understand  in  1,  that  the  joint  capital,  $700,  is  the  basis 
upon  which  the  profits  are  distributed,  they  will  have  no  diflS- 
culty  in  understanding  that  B  is  entitled  to  fJJ  of  $182,  and 
that  C  is  entitled  to  fJ-J  of  $182.  Cancellation  should  be  em- 
ployed.    See  Art.  1121,  No.  5. 

There  is  no  need  in  this  connection  of  discussing  the  subject 
of  business  partnerships  that  are  continued  for  a  year  or  longer. 
The  division  of  profits  in  these  cases  is  the  subject  of  a  special 


NOTES   ON   CHAPTER   TWELVE  133 

agreement,  and  is  rarely  made  solely  on  the  basis  of  the  amounts 
invested.  A  yearly  gain  of  $4000  made  in  regular  business  by 
two  partners,  one  of  whom  invested  $1000,  and  the  other  $3000, 
might  be  divided  in  various  equitable  ways.  The  partner  invest- 
ing the  larger  sum,  might  first  take  out  $  120  as  interest  on  his 
excess  of  capital ;  and  the  remaining  $  3880  might  be  equally 
divided,  giving  one  of  them  $1940,  and  the  other  $2060. 
Another  arrangement  might  permit  each  partner  to  withdraw 
a  fixed  sum  for  services,  say  $1000,  leaving  $2000  to  be  divided 
on  the  basis  of  1  to  3.  This  would  make  the  shares  ($1000 
+  $500)  and  ($1000 +  $1500),  or  $1500  and  $2500. 
10.    Cases  of  this  kind  are  found  only  in  the  books. 

979.  The  cost  of  the  first  item  is  given.  The  second  item 
contains  451  sq.  ft. ;  the  third  contains  (4  X  42  X  7-J-)  sq.  ft.,  or 
1204  sq.  ft. ;  the  fourth  contains  (3  X  43  X  7|)  sq.  ft.,  or  989 
sq.  ft.;  the  total  of  the  three  items  being  2644  sq.  ft.  The 
cost  @  lOd.  is 

£110-    3-4 
First  item,  24-    7-8 

£134-11 
Less  2^%  (^V).  3-    7-3^  + 

Ans.  £  131  -    3  -  8| 

The  fraction  of  a  penny  in  the  discount  is  ■^,  which  is  nearly 
1  farthing,  written  \d. 

980.  German  currency  being  a  decimal  one,  the  bill  is  com- 
puted in  the  ordinary  way.  The  duty  is  found  by  reducing  the 
marks  to  dollars  by  multiplying  by  $.238,  and  taking  35%  of 
the  result. 

981.  Too  much  stress  cannot  be  laid  upon  the  importance  of 
requiring  children  to  estimate  the  probable  answer  to  every 
"  written  "  problem  before  placing  a  figure  on  paper.  The  mere 
drill  on  the  "  approximations  "  found  in  the  text-book  is  of  com- 


134  MANUAL   FOR   TEACHERS 

paratively  little  value,  unless  it  leads  pupils  to  the  employment 
of  this  device  throughout  all  of  their  work.  Besides  preventing 
a  pupil  from  making  a  very  serious  mistake  in  an  ordinary  com- 
putation, the  habit  of  careful  reading  that  is  necessarily  formed 
by  the  scholar  that  is  not  satisfied  with  a  simple  guess,  will 
tend  to  make  his  methods  simpler  and  more  accurate.  He  will 
learn  to  apply  to  the  apparently  more  difficult  numbers  of  the 
written  problem  the  processes  employed  in  solving  the  compara- 
tively simple  "  mental  "  questions. 

While  all  of  the  pupils  should  not  be  expected  to  give  the 
same  answer  to  each  of  these  examples,  they  should  gradually 
approach  more  and  more  closely  to  the  correct  result. 

1.  480  is  what  per  cent  of  960  ? 

2.  52  bu.  is  about  3  times  (17  bu.  37  lb.). 

3.  500  cu.  ft.  -^  128  cu.  ft.     Less  than  4  cords. 

4.  120  cu.  ft.  -^  about  1^  cu.  ft. 

5.  1500  sq.  rd.  -^  160  sq.  rd.     Less  than  10  acres. 

6.  A  little  less  than  70  X  70. 

7.  (6  X  4  X  5)  X  about  7|-. 

8.  64  -^  ^-^,  or  64.3  -^  ■^.     More  than  643. 

9.  About  £  200  @  f  4.80  to  £. 

10.    About  4  marks  to  $1.     Over  400  marks. 

982.  The  teacher  that  is  allowed  nny  discretion  should  omit 
all  problems  relating  to  Bonds  and  Stocks.  The  average  gram- 
mar-school pupil  cannot  be  made  to  understand  the  subject 
without  the  expenditure  of  more  time  and  energy  than  should 
be  given  to  a  topic  that  he  will  learn  by  himself  when  he  grows 
older  if  he  gets  the  proper  foundation. 

If,  however,  the  course  of  study  requires  that  this  topic  be 
taken  up,  the  teacher  should  aim  to  interest  the  pupils  by  making 
some  local  corporation  the  basis  of  the  work.  A  certificate  of 
stock  should  be  obtained,  or  at  least  a  copy  made,  which  might 
be  placed  upon  the  blackboard. 


NOTES   ON   CHAPTER   TWELVE  135 

The  scholars  should  be  led  to  see  that  large  undertakings,  such 
as  the  construction  of  a  railroad,  the  building  of  water-works, 
and  the  like,  require  more  money  than  any  individual  might 
have,  or  would  care  to  risk.  It  then  becomes  necessary  to 
interest  a  number  of  persons  that  will  be  willing  to  invest  more 
or  less  money  in  the  new  enterprise.  It  is  found,  for  instance, 
that  $  50000  will  be  needed  to  build  and  equip  the  street  rail- 
road mentioned  in  1.  The  projectors  divide  this  amount  into  500 
shares,  each  of  which  represents  a  one-five-hundredth  interest 
in  the  profits.  It  may  happen  sometimes  that  it  is  considered 
advisable  to  interest  people  of  small  means,  and  who  are  unwill- 
ing to  take  a  $  100  share.  In  these  cases  the  original  (par)  value 
of  the  shares  may  be  fixed  as  low  as  $10  each.  When  the  par 
value  is  not  given,  it  is  understood  to  be  $100. 

In  the  distribution  of  profits,  the  owner  of  10  shares  is  entitled 
to  3^4  of  $2000,  or  $40. 

2.  These  profits  are  generally  distributed  annually,  semi- 
annually, or  quarterly.  Before  the  time  comes  for  "  declaring  " 
the  dividend,  the  directors  of  the  company  meet  and  determine 
how  much  money  shall  be  thus  distributed.  It  may  be  con- 
sidered advantageous  to  reserve  a  portion  of  the  profits  for  the 
purchase  of  new  cars,  or  for  the  extension  of  the  road,  etc. ;  so 
that  the  amount  distributed  at  any  time  does  not  necessarily 
include  all  that  has  been  gained.  The  dividend  is  generally 
announced  as  a  per  cent  of  the  capital,  which  in  this  case  is 
$50000;  so  that  the  semi-annual  dividend  is  4%,  equal  to  8% 
per  year. 

3.  The  owner  of  a  $100  share,  on  which  he  receives  $8  per 
year,  is  not  likely  to  be  willing  to  sell  it  for  $100  if  he  can 
obtain  only  $4  per  year  interest  on  that  sum  of  money  deposited 
in  a  savings  bank.  The  person  desirous  of  obtaining  stock  after 
it  is  reasonably  certain  that  the  railroad  is  going  to  prove  suc- 
cessful, will  have  to  pay  more  than  $100  per  share.  Mr.  H.  pays 
$150  per  share,  or  150%  of  the  par  value. 


136  MANUAL   FOR  TEACHERS 

4.  Mr.  H.  receives  4%  dividend  on  $3000,  or  $120.  From 
the  savings  bank  he  would  obtain  2%  on  $4500,  or  $90. 

5.  The  $120  semi-annual  dividend  is  2|%  on  the  $4500 
invested,  equivalent  to  5^%  per  year. 

6.  The  words  "  per  cent  "  are  not  used  in  stating  the  price  of 
stock. 

A  $100  share  at  164^%  is  worth  $164|-;  30  shares  are  worth 
$1641x30. 

7.  Assuming  the  par  value  of  each  to  be  $  100,  the  first  pays 
$6  per  year  on  $150,  or  4%  ;  the  second  pays  $7  per  year  on 
$175,  or  4%. 

The  par  value  of  $100  is  assumed  for  convenience;  a  par 
value  of  $50  would  make  the  cost  of  a  share  of  gas  stock  150% 
of  $50,  or  $75,  and  its  annual  dividend  would  be  6%  of  $50,  or 
$3,  the  rate  on  the  amount  invested  being  4%. 

8.  The  buyer  of  stock  at  125  wishes  to  receive  4%  of  $125, 
or  $5.  As  the  dividend  is  based  on  the  par  value  ($100),  the 
rate  must  be  5%  per  year,  or  2^%  semi-annually. 

9.  [93|-%  of  ($50  X  17)]  +  [102f  %  of  ($10  X  143)]. 

10.  A  person  that  desires  to  buy  stocks  is  not  always  likely 
to  know  where  he  can  find  any  for  sale;  so  he  goes  to  a  stock- 
broker, who  makes  a  business  of  buying  and  selling  stocks  on 
commission.  This  commission  is  a  small  per  cent  of  the  par 
value,  the  charge  of  ^%  for  buying  or  selling  a  share  of  the  par 
value  of  $100  being  12^^,  whether  the  actual  value  of  the  stock 
be  $150  or  $50. 

This  broker  receives,  therefore,  \%  of 

[($50x17)  + ($10x143)]. 

11.  A  bond  is  a  note  issued  by  a  corporation,  and  is  generally 
secured  by  a  mortgage  on  its  property.  It  is  a  much  larger 
document  than  the  note  of  an  individual,  and  frequently  contains 
at  the  bottom  a  number  of  "  coupons,"  one  for  each  half-year's 
interest,  upon  which  is  engraved  the  date  when  due  and  the  sum 


NOTES   ON    CHAPTER   TWELVE  137 

payable.  A  10  years'  U.  S.  4%  coupon  bond  for  $  1000  would 
have  20  coupons,  each  worth  $20  when  due.  At  the  expiration 
of  each  6  months,  the  holder  of  the  bond  cuts  off  the  proper 
coupon  and  presents  it  for  payment.  A  "  registered  "  bond  con- 
tains no  coupons,  a  check  for  the  interest  being  mailed  to  the 
owner. 

Although  the  railroad  company  in  this  example  receives  only 
$95  for  each  $100  bond,  it  promises  to  pay  $4  interest  per  year, 
and  $  100  at  the  end  of  20  years. 

In  considering  the  rate  of  interest  received  by  the  owner  of 
such  a  bond,  it  is  not  customary  to  complicate  the  example  too 
much  by  requiring  the  pupils  to  take  into  account  the  additional 
$5  above  the  cost  received  when  the  bond  is  redeemed  at  the  end 
of  20  years,  although  buyers  of  bonds  include  it  in  their  calcula- 
tions. For  our  purpose,  at  present,  it  will  be  sufficient  to  assume 
that  the  purchaser  of  one  of  these  $  100  bonds  receives  $  4  on 
each  $95  invested,  the  rate  being  400^95,  or  4^^%. 

12.  Omitting  the  question  of  redemption,  at  which  time  the 
purchaser  for  $116.50  would  receive  only  $100,  the  rate  is 
400-116^,  or  3ifi%. 

The  holder  of  a  U.  S.  bond  knows  that  the  face  value  of  the 
bond  will  be  paid  in  full  at  maturity,  and  that  the  interest  pay- 
ments will  be  made  on  the  dates  when  due ;  in  the  case  of  the 
bond  of  a  railroad,  or  the  like,  there  is  always  the  possibility 
that  something  may  occur  to  prevent  the  company  from  meeting 
its  obligations. 

.  13.  The  rate  of  income  from  stocks  may  vary  at  each  dividend 
period,  depending  upon  the  amount  of  business  done,  etc. ;  the 
rate  of  income  from  bonds  is  fixed  as  stated  on  their  face. 

Bonds  are  redeemed  at  the  time  specified ;  there  is  no  reason 
why  a  successful  company  should  sell  out  and  divide  the  proceeds 
among  its  stockholders.  When,  however,  the  property  of  a  cor- 
poration is  sold,  the  claims  of  bondholders  and  all  other  obliga- 
tions must  be  satisfied  before  the  stockholders  receive  anything. 


138  MANUAL   FOR   TEACHERS 

14.  The  stock-broker's  fee  is  called  brokerage,  or  commission. 

15.  A  cotton-mill  obtains  material  through  a  cotton  '*  factor  "  ; 
property  is  purchased  through  a  real  estate  agent ;  a  grocer  may- 
buy  butter,  eggs,  etc.,  from  a  commission  merchant,  the  seller  in 
each  case  remitting  the  amount  received  to  the  owner  after 
deducting  his  fee,  or  commission. 

16.  The  hose  in  insurance  is  the  sum  for  which  the  property 
is  insured;  in  taxes,  it  is  the  assessed  value  of  property;  in 
brokerage,  it  is  the  par  value  of  stocks  or  bonds ;  in  commission, 
it  is  the  sum  for  which  goods  are  bought  or  sold  ;  the  principal  is 
the  base  upon  which  interest  is  calculated ;  the  face  of  the  note 
is  the  base  in  bank  discount ;  in  commercial  discount,  the  gross 
price  is  the  base  in  the  first  instance,  the  base  for  each  subsequent 
discount  being  the  successive  remainders  left  after  the  deduction 
of  the  previous  discounts ;  in  stocks  and  bonds,  the  base  is  the 
par  value. 

17.  The  assessed  value  of  property  is  the  value  for  purposes 
of  taxation,  and  is  fixed  annually  by  ofl[icers  chosen  for  this  duty, 
generally  called  assessors. 

18.  2J%  of  assessed  value  (a;)  =  $540;  assessed  value  = 
$24000.     This  is  |  of  the  actual  value,  or  66|%. 

For  various  reasons,  the  assessed  value  is  placed  below  the  sum 
that  would  be  realized  by  the  sale  of  the  property  under  favor- 
able conditions ;  but  care  is  taken  that  all  property  is  assessed 
upon  the  same  basis. 

19.  If  all  property  were  assessed  at  its  actual  value,  the  same 
amount  of  taxes  would  be  produced  by  a  lower  rate.  To  obtain 
$540  taxes  on  property  assessed  at  $36000,  the  rate  would  be 

21.  1674  ft.  =  558  yd. ;  558  yd.  ^  5^  yd.  =  1116  half-yd. 
^11  half-yd.,  which  gives  101  rd.  and  5  half-yd.  =  101  rd. 
2^  yd.  -  101  rd.  2  yd.  1  ft.  6  in. 

22.  $ 8575  H-  $ 245  =r  35  =  number  of  shares.  Quarterly  divi- 
dend =  2^%  of  ($100  X  35). 


NOTES   ON   CHAPTER   TWELVE  139 

983.     27.    Each  figure  of  the  product  is  written  two  places  to 
the  right  of  the  corresponding  figure  of  the  multiplicand. 

Principal,     $375. 


'0 


11.25 


$386.25       Amount  I  yr. 
3%  11.5875 

$397.8375  Amount  1  yr. 
%%         11.9351 


$409.7726  Amount  1-J- yr. 
12.2931 


$422.0657  Amount  2  yr. 
1%  )        4.2206 
\%  i        2.1103 

$428.3966  Amount  2  yr.  3  mo. 
Principal,     375. 

$53.3966  Interest  2  yr.  3  mo. 
Am.  $428.40,  amount;  and  $53.40,  interest. 

It  is  not  necessary  throughout  the  work  to  carry  the  multi- 
plication beyond  four  places  of  decimals. 

985.  6.  After  deducting  the  first  25%,  or  i,  f  of  list  price 
remains ;  a  second  discount  of  \  of  this  remainder  leaves  }  of 
this  remainder,  or  J  of  f  of  list  price,  or  -^^  of  list  price  =  90^. 
List  price  =  90^-^3^  =  90^Xi/  =  $  1.60. 

7.  All  together  can  do  (i  +  -g-  +  i)  of  the  work  in  1  hr. 

8.  Selling  price,  $60  ==  f  of  value  of  cow  ;  \  of  value  =  $60 
-^3  =  $20,  the  loss. 

9.  Selling  price,  $60  =  |  of  value  of  cow  ;  J  of  value  =  $  60 
-^5  =  $12,  the  gain. 

986.  6.  The  wrong  weights  are  ^5^,  or  |^  of  correct  weights, 
so  that  the  customer  receives  for  $352,  W  of  this  amount,  the 


140  MANUAL   FOR   TEACHERS 

gain  to  the  grocer  being  -^^  of  $352,  or  $16.50.  By  selling 
16J  oz.  to  the  pound,  the  grocer  gives  |-|  of  the  proper  amount, 
his  loss  being  -^  of  $  320,  or  $  10.  The  net  gain  is  $  16.50  -  $  10 
=  $6.50.  Ans. 

7.  Cost  of  alcohol,  $2.50  X  42  =  $105  ;  3  yr.  interest  on  $105 
=  $18.90.  Amount  to  be  realized,  $105 +  $18.90  =  $123.90. 
Number  of  gallons  to  be  sold,  42  —  7  =  35.  Selling  price  per 
gallon,  $123.90  -^  35  =  $3.54. 

8.  £4500  =  $4.85x  4500  =  $21825.  Income  on  consols  at 
3%  =  $654.75.  Selling  price  of  consols,  $21825  X. 96;  value 
of  U.  S.  bonds,  $21825  X  .96  h-  108.  Canceling,  we  obtain 
$19400;  6%  of  which  gives  the  income  on  bonds  =  $1164. 
Difference  =  $1164  -  $654.75  =  $509.25.  Ans. 

9.  Number  of  cubic  feet  =.  9|  X  9^  X  6f  =  609 ;  weighing 
609000  oz. ;  etc. 

988.  11.  Agent  collected  80%  of  $4500  =  $3600;  on  this, 
his  commission  at  7^%  is  $270;  making  the  amount  to  be 
given  me  =  $  3600  -  $  270. 

13.   a;  X -I- X  I  or  5^  =  15.12. 
100      2        40 

Find  the  discount  for  17  da.,  the  time  from  June  20  to  July  7. 

15.  Ignoring  the  price  —  if  he  could  buy  80  lb.  with  81^%  of 
his  money,  he  could  buy  with  100%,  (80  lb.  h-  81^)  X  100. 

17.  Let   a:  =  cost  of  each  cow;    6 a;  =  cost  of  6  cows;    com- 

S      r   n         18a:  .1  •    •  c      1  18a; 

mission  =  — —  of  6  a;  =  — - ;    cost   and   commission  =  6  a;  +  — — 

100  100  100 

=  525.30. 

18.  The  note  is  discounted  35  da.  after  it  is  made,  so  that  it 
has  (93  —  35)  da.  to  run,  or  58  da.  [Without  grace,  55  da.] 
The  interest  for  a  year  is  $36,  which  is  10  cents  a  day,  or  $5.80 
for  58  days ;  etc. 

19.  If  the  selling  price  (regardless  of  its  amount)  is  six-fifths 
of  the  cost,  the  gain  is  -J-  of  cost,  or  20%. 


NOTES   ON    CHAPTER   TWELVE  141 

989.  6.  The  walls  contain  [(20  +  15  +  20  +  15)  X  10]  sq.  ft. ; 
the  ceiling  contains  (20  X  15)  sq.  ft.  Dividing  by  \^,  the  width 
in  feet  of  the  paper,  gives  the  number  of  feet  of  paper  required, 
which  is  then  reduced  to  yards. 

991.  7.    Dividend  is  3^%  of  ($9562.50^  1.27^). 

993.  Mr.  Smith  wishes  to  pay  Mr.  Thompson  the  exact 
amount  of  his  bill.  A  check  on  a  Memphis  bank  for  $3475.86 
would  not  be  sufficient,  as  Mr.  Thompson  would  have  to  pay  a 
New  York  bank  for  collecting  the  check  in  Memphis.  As  the 
charge  may  not  always  be  the  same,  Mr.  Smith  cannot  know 
how  much  to  add  to  the  amount  of  his  bill  to  cover  this  expense. 
From  a  banker  that  has  an  account  in  a  New  York  bank,  he 
can  obtain  a  draft,  payable  in  that  city,  for  the  exact  amount,  by 
giving  the  Memphis  banker  $3475.86  +  $5.21,  or  $3481.07. 

Exchange  is  at  a  premium  when  the  cost  of  a  sight  draft  is 
greater  than  its  face ;  it  is  at  a  discouiit  when  the  cost  of  a  siglit 
draft  is  less  than  its  face. 

Bank  Check. 


^«- 


.    Quogue,  JSr.Y.,  fayyv.  S,  189  6.       Jfo.  /^^i'. 

SHINNECOCK  NATIONAL  BANK. 

Pay  to  the  order  of    £&w-ia.  /C.    3^k^L1tow^,     <^6Sf^^ 

'^i/xXA^-&v^At ^ — .^.^ ^jQQ  Dollars 

HOWELL  &  PENNIMAN. 


An  examination  of  the  above  check  will  show  wherein  it  differs 
in  form  from  the  draft.  A  draft  may  be  made  payable  at  a 
future  time,  whereas  a  check  is  always  payable  on  presentation. 


142  MANUAL   FOR  TEACHERS 

8.  A%=TATr;^-|^-=632.18.  9.   ^+^=1000. 

10.  $339.66  -  (2%  of  $339.66)  =  sum  remaining  for  the  pur- 
chase of  dry-goods,  etc.,  and  the  commission.  Dividing  this  by 
1.02  gives  the  cost  of  the  goods. 

Another  method  of  solving  the  foregoing  is  to  indicate  the 
money  remaining  as  $339.66  X  .98.  Using  1.02  as  a  divisor,  and 
canceling,  gives  the  result. 

339.66  X  98 

102 

994.2.  Noon  Monday  to  6  p.m.  Thursday  =  78  hr.  The 
loss  in  time  =  35  sec.  X  78  =  45  min.  30  sec.  The  time  shown  is 
6  hr.  —  45  min.  30  sec.  =  5  hr.  14  min.  30  sec,  or  14^  min.  past  5. 
4.  The  number  of  rows,  2  ft.  apart  in  a  space  36  ft.  —  4  ft., 
is  (32-f- 2)-f  1  =  17.  The  number  of  plants,  16  in.  apart  in  a 
row  60  ft.  -  2|  ft.,  or  57|  ft.,  in  length,  is  (57^^  IJ)  +  1  =  44. 
Total  number  of  plants  =  44  X  17  ==  748. 

If  the  rows  run  crosswise  there  will  be  29  of  them,  each  con- 
taining 26  plants. 

6.   Number  of  revolutions  =  14  mi.  -^  13  ft.  4  in. 

995.  8.  Length  of  a  degree  on  the  equator  =  25000  mi.  -^-  360. 
20°  will  measure  (25000  ^  360)  X  20.     Cancel. 

9.  The  circumference  =  18  ft.  X  3.1416.  Divide  by  360. 
Cancel. 

10.  The  difference  is  20°,  and  the  distance  will  be  about  one- 
half  that  found  in  8. 

The  teacher  should  remember  that  the  shortest  distance 
between  these  two  places  is  not  measured  on  the  parallel  of  60°. 
The  shortest  distance  between  two  points  on  a  sphere  is  measured 
by  the  arc  of  a  great  circle  joining  the  points,  and  the  20°  are  -^ 
of  a  small  circle. 

11.  46°22'30"  =  46f°.     The  number  of  miles  =  69J  x  46f 

12.  The  approximate  length  of  the  45th  parallel  is  25000  mi. 
X  .7071 ;  the  length  of  a  degree  on  this  parallel  =  (25000  mi. 


NOTES   ON   CHAPTER   TWELVE  143 

X  .7071)  -^  360;  multiplying  by  22^  gives  the  required  distance. 
Cancel. 

996.  Time  drafts  are  so  little  used  that  it  is  scarcely  worth 
while  to  spend  much  time  on  their  study. 

A  sight  draft  being  payable  on  presentation  (except  in  those 
states  allowing  days  of  grace),  there  is  no  need  of  formal  accept- 
ance. Acceptance  is  necessary  in  the  case  of  time  drafts,  as  they 
are  not  payable  until  the  specified  time  after  this  acceptance. 

The  acceptance  of  a  draft  makes  the  person  or  corporation 
accepting  it  liable  to  its  owner  for  the  amount,  a  draft  being 
transferable  by  endorsement  just  as  a  check  or  a  note. 

997.  In  calculating  the  cost  of  a  sight  draft,  days  of  grace  — 
even  when  allowed  —  do  not  enter  into  the  result,  this  being 
included  in  the  rate  charged.  Time  drafts  are  allowed  days 
of  grace,  except  in  the  states  given  in  the  Appendix,  Art.  1305. 
The  number  of  states  in  which  days  of  grace  are  no  longer 
allowed,  increases  yearly,  there  being  no  good  reason  for  promising 
to  pay  in  60  da.  when  the  intention  of  the  signer  is  to  take  63  da. 

1000.  1.  Although  days  of  grace  are  not  allowed  in  Cali- 
fornia, the  pupils  of  other  states  should  not  be  expected  to  know 
this.  In  states  that  grant  days  of  grace,  they  should  be  allowed 
in  every  note  or  time  draft,  no  matter  where  payable ;  while  in 
the  other  states,  pupils  should  be  taught  not  to  employ  them 
in  any  case. 

The  premium  on  the  draft  =  1 1.75  X  .840  =  1 1.47.  The 
interest  (with  days  of  grace)  =  $840  X  yf^  X  ^\  =  $13.02. 
The  cost  of  the  draft  =  $840 -f$  1.47- $13.02  =$828.45.  Ans. 

Or,  without  days  of  grace  : 

$840  X  yf^X -jV^  =  $  12.60,  the  cost  being  $840 +  $1.47 
-$12.60  =  $828.87.     Ans. 

Some  teachers  prefer  to  find  the  cost  of  a  draft  for  $1  at  the 
given  premium  —  in  this  case,  $1.00175  ;  from  which  is  deducted 
the  interest  on  $1  for  93  da.,  or  $.0155;  making  the  cost  of  a 


144  MANUAL    FOR   TEACHERS 

90-day  draft  for  $1,  $1.00175  -$.0155  =$.98625.  Multiplying 
this  by  840  gives  $825.45,  the  cost  of  a  draft  for  $840.  This 
method  is  not  so  much  shorter  as  to  make  it  advisable  to  use  it. 

2.  The  discount  =  ^%  of  $400  =  ^  of  $400=  50^.  The 
interest  for  33  da.  =  $400  X  yf^  X  ^^  =  $2.20.  Cost  =  $400 
-$.50 -$2.20  =  $397.30.  Ans. 

Note.  —  The  word  "interest"  is  used  instead  of  "bank  discount,"  to 
avoid  the  confusion  arising  from  the  use  of  "discount"  with  two  meanings 
in  the  same  example. 

6.  The  six  remaining  examples  should  be  omitted  by  pupils 
that  do  not  use  algebraic  methods  of  solution.  Scholars  that 
have  readily  worked  the  first  four  examples  will  find  no  great 
difficulty  in  solving  5-10  by  means  of  the  equation. 

The  premium  is  $^  per  $1000,  or  :f7nnr  ^^  ^^®  ^^^®  ^^  ^^® 

draft,  X.     Premium  = The  interest  on  x  dollars  for  63  da. 

4000 

at  6%  =a;X— —  X— —  =  -— — •     Adding   the   premium   to  the 
^  100     360     2000  &  f 

face  of  the  draft,  and  deducting   the   interest,  gives  a:-}-  .   ^^ 

21 X  .  4000 

—  as  the   cost   of    the   draft.     This  may  be  changed  to 

4000  x  +  x-^2x_  3959  x     . 

4000  4000  '      '^' 

6.   i%  of  $1200  =  $1.50;   interest  for  (a: +  3)  da.  =  1200  X 

— —  X^^t"  =^~l  ;  cost  of  the  draft  (with  days  of  grace), 
100      360  5  V  J  6       /• 

1200     U      ^  +  ^-l^QQQ     1^     2^      6  _  11979 -2  a:     j^^ 
*         5  10        10      10      10  10        ■  ■ 

Note.  —  Unless  the  pupil  has  studied  algebraic  subtraction  in  Chap.  XV., 
he  may  make  a  mistake  in  deducting  ^  "^    ,  by  failing  to  change  the  sign 

of  the  second  term.     By  writing  ^       »  -  +  -,  he  may  see  more  clearly  that 

5      5     5 

the  cost  of  the  draft  is  1200  —  1^  —  -  — ,  etc.     Without  days  of  grace,  cost 

-1200-11--;  etc.  ^     ^ 

5 


NOTES   ON    CHAPTER    TWELVE  145 

-   1001.  1.    15°x3i  =  50°.  Ans. 

2.  (61  -^  15)  hr.  =^  4^5  lir.  =  4  hr.  4  min.  Ans. 

3.  Difference  in  time  —  -J^  hr.  =  5  hr.  London  time  =  1  p.m. 
+  5  hr.  =  6  P.M.  Ans. 

4.  2  P.M.  —  5  hr.  =  9  a.m.     Ans. 

5.  Vienna  is  -f-l  hr.  later  =  1-|  hr.  ==  1  hr.  40  min.  Time  at 
Vienna  40  min.  after  1  p.m.,  or  20  min.  to  2  p.m.  Ans. 

6.  3  hr.  40  min.  =  3|  hr.  Difference  in  longitude  =  15°  X  3| 
=  55°.  Ans. 

7.  Difference  in  longitude  =  75°  +  30°  -=  105°.  Ans. 

8.  Philadelphia  time  is  ^^-  hr.  earlier,  or  7  hr.  3  p.m.  — 
7  hr.  =  8  a.m.  Ans. 

9.  Correct  Washington  time  is  j\  hr.,  or  8  min.  earlier  than 
standard  time. 

10.  A  town  in  84°  west  longitude  is  6°  east  of  90°,  so  that  its 
correct  time  is  -^^  hr.,  or  24  min.,  later.    Time,  12  :  24  p.m.  Ans. 

1002.  1.  Longitude  difference  =  15°  X  3|-f .  The  pupil  should 
see  that  15  X  |^  =  44  ^  4  =  11  ;  so  that  15°  X  3*^  =  45°  + 11° 
=  56°.  Ans. 

2.  15)37  hr.  18  min. 

2  hr.  29  min.  12  sec. 

At  1  hr.  to  a  degree,  the  difference  in  time  would  be  37  hr. 
18  min. ;  as  it  requires  15°  to  make  an  hour's  difference,  dividing 
37  hr.  18  min.  by  15  gives  the  result. 

Shorter  methods  should  be  deferred  for  the  present.  Using 
multiplication  to  obtain  the  difference  in  degrees,  and  division  to 
obtain  the  difference  in  time,  is  more  easily  understood  by 
beginners. 

3.  Time  difference  =  y^-g-  of  87  hr.  35  min.  earlier  at  Chicago, 
because  it  is  west  of  Greenwich.  Standard  Chicago  time  is  the 
time  of  90°,  or  90  hr.  -;-  15  =  6  hr.  earlier  than  Greenwich. 
Standard  time  =  1  P.M.  —  6  hr.  =  7  a.m.  Ans. 


146  MANUAL    FOR  TEACHERS 

4.  Vessel's  time  is  2^  hr.  earlier,  showing  that  the  vessel  is 
15°  X  2i,  or  37^°  west  of  Greenwich.     Am.  37°  30'. 

6.  Time  difference  =  H  hr.  Longitude  difference  ==  15°  X 
1-^  =  22J°*  The  latter  place,  having  the  later  time,  is  the  more 
easterly ;  so  that  its  longitude  is  22 J°  east  of  11°  east,  or  33° 
30'  east.  Am. 

7.  3  da.  12  hr.  17  min.  =  84|J  hr. ;  3313.5  h-  84fJ  =  number 
of  miles  per  hour. 

11.  12^  (ft.)  X  3f  (ft.)  X  X  (ft.)  =  730^;%  =  730^  (cu.  ft.) ; 

=  1^=  15^^.     Am.  15f^  ft.  -  15  ft.  7  in. 

12.  48%  =  237  bu.  3  pk. 

+  4%  =    19  bu.  3  pk.  2  qt.  -j^  of  48% 

Remainder,      52%  =  257  bu.  2  pk.  2  qt.  Am. 
14.    Number  of  degrees  -=  (34  X  24)  -f-  48.96. 

1003.  5.    84  half-dollars -84  cents  ==$42.00 -$.84. 

9.  After  taking  ^,  |-  are  left;  when  ^  of  the  remainder  is 
taken,  f  of  remainder  are  left,  or  -J  of  -f  =  -J  =  4  gal. ;  etc. 

1004.  5.    12  men  working  8  hr.  daily  build  90  rd.  in  15  da.; 

7  men  working  10  hr.  daily  build  70  rd.  in  ?  days. 

15  da.  X  12  X  8  X  70 
90  X  7  X  10 

9.    72  (in.)  X  48  (in.)  X  x  (in.)  =  2150.4  (cu.  in.)  X  75. 

16.  a;  +  (a: +15) +  (5: +15  4- 27)  =  320. 

17.  .64  bu.  =  4  pk.  X  .64  =  2.56  pk. ;  .56  pk.  =  8  qt.  x  .56  = 
4.48  qt. ;  3.64  bu.  =  3  bu.  2  pk.  4.48  qt. ;  -^^  bu.  =  4  pk.  X  A 
n=  I  pk.  =  2i  pk.  =  2  pk.  2  qt. ;  3  bu.  2  pk.  4.48  qt.  +  2  pk. 
2  qt.  +  1  bu.  3  pk.  6.52  qt.  =  6  bu.  5  qt. ;  10  bu.  —  6  bu.  5  qt. 
=  3  bu.  3  pk.  3  qt.  Am. 

18.  Each  step  takes  7^  in.  +  10  in.  ^  17^  in.  =  ^  yd. 
Co8t  =  90^X-y^X  18..    Cancel. 

24.    a:  +  (a: +1211)  =  9891. 


NOTES   ON   CHAPTER   TWELVE  147 

1005.  Formerly,  bills  of  exchange  were  issued  to  purchasers 
in  sets  of  three  bills,  two  of  which  were  sent  by  different  steamers 
to  the  foreign  payee,  who  presented  for  payment  or  acceptance 
the  one  that  reached  him  first.  The  third  bill  was  retained  by 
the  purchaser,  to  be  sent  in  case  both  of  the  others  failed  to 
reach  their  destination.  At  present,  only  two  bills  of  a  set  are 
issued.     The  second  will  read  as  follows : 

Exchange  for  £180  17s.  6c?.  New  York,  Dec.  14,  1895. 

Sixty  days  after  sight  of  this  Second  of  Exchange  (First 
unpaid)  pay  to  the  order  of  John  W.  Moran  &  Bro.,  One  Hun- 
dred Eighty  pounds  sterling,  seventeen  shillings,  six  pence. 

Value  received,  and  charge  the  same  to  account  of 

To  James  Lennon  &  Co.,  )  -r»  r^  o    oi 

London.  i  PeTER   CoMERFORD   &   SON. 


No.  39. 

1.  No  deduction  for  interest  is  made  for  the  60  da.,  the  quo- 
tation giving  the  price  per  pound  for  60-day  bills. 

The  method  given  in  the  text-book  is  a  form  of  the  aliquot 
part  method  used  in  calculating  interest. 

1007.     3.    Cost  in  dollars -1000  ^5.1625. 

4.  Cost  in  dollars  =  1874.35  X  .9525  ^  4. 

5.  Number  of  marks  =  1000  ^  (.955  ^  4)  =  4000  -^  .955. 

6.  Number  of  francs  ==  1637.5  X  5.185. 

8.  £437    5s.     lOd'. 
Less  4%,  or  ^,      17     9s.     10c?. 

£419  16s. 
Cost  =  $4,885  X  419if  =  $4,885  X  419.8. 

9.  18pcs.,44m.  each,  or  792  m. 

@  fr.  25  =  fr.  19800 


Less  7|-%, 

1485 

fr.  18315 

3  pes.  50  m.  each,  or  150  m.  @  fr.  20, 

fr.3000 

Less    5%, 

150 

2850 

Packing  charges, 

60.50 
fr.  21225.50 

Cost  of  bill  =  19^^  X  21225.5  =  $  4138.97.  Am. 


148  MANUAL   FOR  TEACHERS 

10.  M.  3598.60 

Less  10%,  359.86 

M.  3238.74 
Less    5%,  161.94 

M.  3076.80 
Less2|%,  76.92 

M.  2999.88 
Freight,  165  kilos  @  M.  4.80,  792.00 

M.  3791.88 
95J^X  3791.88 ^4 -$908.87.  Am, 


TJNITEKSITT 


XVI 

NOTES  ON   CHAPTER  THIRTEEN 

1008.  The  word  "endorsement"  means  something  that  is 
written  on  the  back  of  a  document.  As  applied  to  notes,  checks, 
drafts,  etc.,  it  generally  means  the  signature  of  the  person  in 
whose  favor  the  note,  check,  etc.,  is  made  out,  which  is  written 
on  the  back  in  order  to  transfer  the  ownership.  If  the  payee  of 
the  following  note  sells  it  to  William  Simms,  he  writes  his  name 
on  the  back,  as  shown  below. 

AccoTiNK,  Va.,  March  4,  1897. 

Four  months  after  date,  I  promise  to  pay  to  the  order  of 
James  McWilliams,  Two  Hundred  Dollars,  value  received,  at  the 
Pohick  National  Bank. 

I2OO1-V7.  Victor  Struder. 


The  effect  of  the  "  endorsement  in  blank  "  is  to  make  it  pay- 
able to  any  holder;  the  "  endorsement  in  full"   transfers  it  to 
I      William  Simms,  who  may  transfer  it  to  another  by  either  kind 
of  endorsement. 

149 


150  MANUAL   FOR  TEACHERS 

Besides  transferring  ownership  in  a  note,  the  effect  of  an 
endorsement  is  generally  to  bind  the  signer  to  pay  the  note,  in 
case  of  default  by  the  maker  or  preceding  endorsers.  This  lia- 
bility is  avoided  if  the  endorser  writes  after  his  name  the  words, 
"  without  recourse." 

The  *'  endorsements  "  mentioned  in  this  chapter  are  a  record 
of  the  payments  received  by  the  holder  of  the  note.  This  is 
usually  kept  on  the  back  of  the  note,  the  date  and  the  amount 
received  being  written  in  each  instance. 

1010.  Although  the  maker  of  a  note  is  generally  supposed  to 
pay  the  interest  at  the  end  of  each  year,  the  U.  S.  Courts,  by 
whom  this  rule  has  been  formulated,  do  not  permit  the  collection 
of  interest  upon  deferred  payments  of  interest. 

This  rule  is  followed  in  all  the  states  except  Connecticut  (see 
Art.  1307)  for  computing  the  amount  due  on  notes  that  do  not 
expressly  provide  for  the  payment  of  interest  annually  (Art.  1172). 
Connecticut  pupils  should  learn  only  their  own  rule ;  in  other 
states,  no  reference  whatever  to  the  Connecticut  rule  should  be 
made. 

See  Art.  1307  for  Connecticut-rule  answers  to  the  partial  pay- 
ment examples  of  this  chapter. 

1013.     6.    Let  100  X  =  cost  of  coal ;  2  a;  =  commission. 
102  X  =  cost  of  coal  +  commission  =  7650. 

1015.  1.  The  man  expended  30%  of  50%  of  f  of  his  money, 
or  ^  X  -J-  X  f  of  it ;  which  was  yf^  of  his  money.  -rJ^a;  =  1-|- 
X  728  =  819 ;  x  =  9100  Two-fifths  of  $9100  equals  the  balance 
in  bank. 

2.  Cost  per  gallon,  $  1.50;  selling  price,  $1.60;  gain  per 
gallon,  10^  on  150)Z^,  or  yV  =  6|%. 

5.    Let  10a;  =  cost,  then  x  will  represent  the  loss  in  one  case 
and  the  gain  in  another,  making  the  selling  prices  11  a;  and  9  a;. 
11  a;  =  99 ;  a;  =  9,  gain  in  dollars. 
9  a;=  99;  a;  =  11,  loss  in  dollars,  making  the  net  loss  $2. 


NOTES   ON    CHAPTER   THIRTEEN  151 

Note.  —  Some  teachers,  wishing  to  avoid  fractions  as  far  as  possible  in 
equations,  assume  x  for  loss  or  gain,  making  10 a;  =  cost;  etc.  Solutions 
of  this  kind  are  given  occasionally  as  a  suggestion  to  be  followed  or  not,  as 
may  seem  most  desirable. 

8.    Arithmetic,  Art.  924,  7  and  8.  ■: 

1016.  As  there  is  no  such  thing  as  "  true  discount,"  it  is  un- 
profitable to  spend  time  upon  it'.  Any  problem  involNring  find- 
ing the  "  present  worth  "  can  be  solved  by  an  intelligent  pupil, 
from  his  previous  work  in  interest. 

1019.  2.   To  \\%  of  I  of  %  25000,  add  $  1. 

3.    Longitude  difference  =  5°  59'  18".     See  Art.  1002,  2. 

10.  Number  of  yards  =  (5616  ^  1.04)  ^  \\.     Art.  1013,  6. 

11.  Term  of  discount  36  (33)  days.  Yearly  interest  is  |30, 
which  is  $3  for  36  da.,  yV  y^'-  $500  -  $3  =  proceeds,  $497. 
Without  grace,  the  term  is  33  days,  the  interest  for  which  time  is 
$2.75.     Proceeds,  $497.25. 

13.  Cost  of  an  acre  =  $21.78  X  5.  Cost  per  square  foot  = 
($21.78x5) -4- (4840x9),  the  divisor  being  the  number  of 
square  feet  in  an  acre.  To  gain  20%  or  \  of  cost,  the  selling 
price  must  be  f  of  cost.  Multiplying  the  foregoing  by  f,  the 
selling  price  per  foot  will  be 

$21.78x5x6 
4840  X  9  X  5  * 
In   getting   the   number  of  square  feet  in  an  acre,  the  pupil  may  use 
160  X  30|  X  9,  unless  he  remembers  that  there  are  4840  sq.  yd.  in  an  acre. 

14.  Specific  duty  (duty  by  weight) -- $i  X  700  X  1^.  Ad- 
valorem  duty  =  30  %  of  $1^  X  700.  The  sum  of  the  two  gives 
the  total  duty. 

1020.  7-9.    See  Arithmetic,  Art.  384. 

1021.  4.    (50  ft.  +  38  ft )  — (7  ft.  +  2  ft.). 

5.   f  of  cost  of  horse  =  $  90.     Cost  =  $80.     A  selling  price  of 
$100  makes  a  gain  of  $20  =^  of  cost  =  25%.  Ans. 


162  MANUAL   FOR   TEACHERS 

6.    Loss  =  ^  of  cost  =  20%.  Ans. 
12.    100%.    15-16.    See  Art.  878.    19.  ^%  of  $700.    20.  $10 
for  60  da.  +  A  ^^  ?10  for  5  da.     21.    200%.     22.    ^  of  $7. 
26.    4A^A=^42-^^3. 

28.  1^  =  1  of  63  =  49;  ^=7;  a:  =  56. 

29.  ^  =  8J  =  ^,  2a:  =  25  ;  a:  =  12^.  Aris.  12  yr.  6  mo. 

o  o 

30.  5  qt.  =  A  pk. ;  I  =  .625  ;  .625  ^  4  =  .156J  =  .15625. 

1022.     1.    $fx5i(yd.)x4(yd.)^li(yd.). 
?ix¥xtxf     Cancel. 

2.  What  sum  in  4  years  at  6%  will  amount  to  $105.71? 
Arithmetic,  Art.  1017. 

^  +  (^XTf^x4)  =  105.71;  a: +  ?^=^  105.71; 

100a; -f  24a:  =10571;  124a;  =  10571;  a;  =  85.25. 

$85.25.  Ans. 

3.  Term  =  27  da.  -}-  3  da.  =  yV  yr. 

a?-(^XTfcXiV)  =  95;  a;-^=95;  etc. 

4.  Problems  of  this  kind  may  be  solved  without  finding  the 
cost;  18^  per  yard  represents  -^  of  cost;  what  price  will  rep- 
resent f  or  U  of  cost  ?  If  A  of  cost  =  18^,  ^ig-  will  equal  2^, 
and  II  will  equal  24^.  Ans. 

5.  24i%=3j4j^;  ^=1372;    a:  =  5600.     The   whole  real 

estate  was  worth  $5600;  the  part  remaining  after  the  sale  of 
$1372  worth  =  $5600 -$1372  =  $4228.  $14000  +  $4228  = 
$18228.  Ans. 

6.  6  times  (5  X  5)  sq.  ft. 

7.  Mr.  Jones  paid  |%  of  |  of  $ 48000  =  ^^^  X  |  X -^ 

1  u      4UU 

~  $  300.  The  company  loses  $  40000  less  the  premium  received, 
$300. 


NOTES   ON    CHAPTER   THIRTEEN  153 

8.  This  may  be  solved  without  finding  the  cost,  although 
many  pupils  will  prefer  the  more  tedious  way.  $  764.40  repre- 
sents 91%  of  cost;  what  per  cent  does  $894.60  represent? 

91^.  X  89460  ^21M„_^  ,061%;  gam  6if.. 

10.  Cancel.     |3|  X  25  X  8  X  8  ^  128. 

11.  Total  cost,$252.50;  loss=$252.50-$141.40-$111.10, 
which  is  44%  of  the  total  cost. 

12.  [14  (yd.)  +  5  +  14  +  5]  X  3. 

14.  Profit$2perton,  Jof  cost.     Cost  =  $198x3. 

15.  Art.  546.  A  bill  is  receipted  by  writing  the  words, 
"  Received  payment"  at  its  foot,  followed  by  the  date  and  the 
name  of  the  seller  : 

^ev  f.  ^. 

If  the  money  is  received  by  a  clerk,  he  writes  his  initials 
underneath,  preceded  by  the  word  "  per  "  or  "  by." 

16.  ^  =  9f ;  9|a;  =  8|;  etc. 

17.  Omit  4^  bu.  I  gave  away  \  and  f  =  it  +  A  =  tt-  T^® 
remainder  =  ^^  =  .26f  =  26f  %. 

21.  $3  per  yd.  =  80%  of  cost;  $?=  115%. 

22.  Number  of  gallons  sold  =  (65  gal.  X  60)  -  80  gal.  = 
3820  gal.    Selling  price  per  gallon  =  f  of  $  1542  ^  3820.    Cancel. 

23.  %  180  =  \  cost  of  one  horse  =  f  cost  of  other. 

24.  Number  of  square  yards  in  walls  =  (6  +  4  +  6-f4)x3; 
in  ceiling,  6x4.     Number  of  cords  =  (18  X  12  X  9)  ^  128. 

25.  Let  X  =  less,  x  -^\=^  greater  ;  x -\- x -{- \  =  ^\. 
'     26.    16  cu.  yd.  is  :f%  of  (10  X  8  X  2)  cu.  yd.  ? 


154  MANUAL    FOR   TEACHERS 

27.  672  yd.  @  $2^  =  $1512.  Discount  without  grace  = 
$1512  XjhXi  =  $17.64.  Profit  is  $1  per  yard  less  discount 
=  $672 -$17.64  =  $654.36.  Ans. 

The  discount  for  3  days'  additional  (grace)  =  -^j^  of  $  17.64  = 
$.88,  making  the  profit  88^  less  than  the  above,  or  $653.48.  Ans. 

28.  40^  X  [(55x600x5^)  ^27].     Cancel. 

1023.  5.  The  circumference  of  the  wheel  =  distance  traveled 
in  1  revolution  =  1  mi.  94  rd.  2  yd.  1  ft.  h-  526  =  6838  ft.  ^  526 
=  13  ft.  =  4  yd.  1  ft. 

6.    The  weight  in  pounds  =  if f »-  X  9|  X  9J  X  6}. 

8.  Rate  of  income  received  on  6%  bonds  =  6 -^- 1.18  ;  rate 
on  4J%  bonds  =  4:^  -. — — .  The  income  being  the  same,  and 
the  same  amount  being  invested,  the  rates  must  be  equal ;  there- 
fore, §52  ==  450  .  600  a;  =  118  X  450  =  53100  ;  x  =  88 J.     Price 

per  $100  =  $88.50. 

9.  The  shrinkage  being  1  lb.  in  10,  he  must  sell  9  lb.  for  the 
cost  of  10  lb.  to  suffer  no  loss.  10  lb.  cost  $1.80  ;  by  charging 
20^  per  pound,  he  receives  the  cost.  To  gain  20%,  he  must  sell 
for  I"  of  20^,  or  24^  per  pound.  Since  he  loses  4%  of  the  amount 
of  sales,  or  ^,  he  receives  only  f|-  of  the  price  charged  per 
pound.  Therefore  to  receive  24^,  he  must  charge  24  ^-^|4 
=  25^  per  pound. 

[(i8^-.A)x|]-M. 

1024.  1.    12xa:  =  20xf. 

2.  This  may  be  solved  by  analysis,  or  the  following  method 
may  be  employed : 

The  solid  contents  of  first  beam  in  cubic  feet  =  16  X  2J  X  f ; 
of  the  second  =  a:  X  3J  X  -J-}.  The  second  weighs  -f-^M  times  the 
first ;  its  contents,  therefore,  =  f f ff  times  the  contents  of  the 
first.  a;X3iX-J^=16x2Jx|X  ff|| ; 

a;  =  (16x2ixtX«tt)^(3ixM) 
=  16  X  I  X  J  X  fMf  X  3^  X  H.     Cancel. 


NOTES   ON   CHAPTER   THIRTEEN  156 

3.  The  carpet  costs  50^  per  foot.  $-1-  X  22-1-  x  15|-^2J 
=  Ans.  Or,  changing  all  dimensions  to  yards  :  $  1-|-  X  7-|  X  5J 
H-  J  =  Ans. 

4.  As  a  sight  example,  some  pupils  may  see  that  the  width 
of  the  large  box  is  double  that  of  each  small  one,  and  its  depth 
is  three  times  that  of  each  small  one,  so  that  with  the  same 
length  as  the  small  one,  it  would  contain  2  X  3,  or  6,  small  ones. 
A  length  twice  as  great  —  8^  ft.  —  is  required  to  enable  it  to 
hold  12  boxes. 

6.    [i  of  (ISixllf)]  sq.ft. 

22  X  14  X  12  X  1728 
2150.4 

Drop  the  decimal,  point  in  the  denominator,  and  annex  a 
cipher  to  one  of  the  numbers  in  the  numerator.     Cancel. 

9.    49  X  44  X  27  ^  231  =  number  of  gallons.     Cancel. 

10.  Solve  at  sight.     7  yd.,  6  yd.,  4  yd. 

11.  Number  of  gallons  =  5i  X  6  X  7  X  1728  ^  231.  Cancel- 
ing, we  obtain  1728  gal.  One  empties  it  in  (1728  -h  9)  min. 
=  192  min. ;  the  other  in  (1728  ^  7)  min.  =  246f  min. ;  both  in 
(1728  ^  16)  min.  =  108  min. 

12.  The  dimensions  of  the  room  are  6  yd.  and  5  yd.,  and  the 
carpet  is  f  yd.  wide.  6  contains  f  an  exact  number  of  times 
(8),  so  that  if  the  carpet  runs  across  the  room  it  will  take  8  strips 
each  5  yd.  long.  As  5  -^  f  =  6|-,  to  lay  the  carpet  lengthwise 
would  require  6  strips,  and  -|  of  a  seventh  strip,  which  would 
have  to  be  cut. 

Carpet  30  in.  wide,  |  yd.,  could  be  laid  lengthwise  without 
splitting  the  breadths,  5-7-|-,  or  6,  strips  being  needed,  each 
6  yd.  long. 

13.  36  yd.  are  needed  to  cover  the  floor;  including  4-J-  yd. 
cut  off  in  matching  the  pattern'^  40-|-  yd.  must  be  bought  at  95^. 
At  10^  per  yard,  the  sewing  and  laying  should  cost  10)^  X  36 
=  $3.60,  but  the  custom  is  to  charge  for  the  number  of  yards 


156  MANUAL   FOR   TEACHERS 

purchased,  40|,  making  $4.05  ;  (5  X  6)  sq.  yd.  or  30  sq.  yd.  @  5^ 
=  $1.50  for  lining.  Total  cost,  $38.47^  +  $4.05 +  $  1.50  = 
$44.02J,  or  $44.03. 

14.  A  strip  I  of  (18  —  15)  or  |  of  (21  —  18)  is  left  uncovered 
on  each  of  the  four  sides,  or  H  ft.  The  area  of  the  uncovered 
space  in  square  feet  =  (21  X  18)  —  (18  X  15). 

15.  The  number  of  square  feet  in  the  walls  =  (18  +  24 -f  18 
4-  24)  X  9.  The  ceiling  contains  (18  X  24)  sq.  ft.  Deduct 
60  sq.  ft.  for  two  doors,  48  sq.  ft.  for  two  windows,  25  sq.  ft.  for 
the  fireplace.  The  total  number  of  feet  around  the  four  walls 
=  18  +  24  +  18+24  =  84  ft.  Baseboard  will  not  be  required 
at  the  doors,  8  ft. ;  nor  at  the  fireplace,  5  ft.  —  a  deduction  of 
13  ft.,  making  71  running  feet  of  baseboard,  1  ft.  wide,  contain- 
ing, therefore,  71  sq.  ft.  The  total  deduction  from  the  area  of 
walls  and  ceiling,  1188  sq.  ft.,  are  60  sq.  ft.  +48  sq.  ft.  +  25  sq.  ft. 
+  71  sq.  ft.  =  204  sq.  ft.,  leaving  1188  sq.  ft.  -  204  sq.  ft.,  or 
984  sq.  ft.,  to  be  plastered. 

16.  The  first  pile  contains  (25  X  20  X  10)  cu.  ft.  and  costs 
$  1400.  1  cu.  ft.  cost  $  1400  ^  (25  X  20  X  10).  Multiplying  by 
(50  X  40  X  20),  the  number  of  cubic  feet  in  the  second  pile,  gives 
the  cost : 

$1400x50x40x20 
25x20x10 

17.  Pupils  that  endeavor  to  solve  a  problem  without  examin- 
ing the  conditions,  will  be  likely  to  assume  that  this  example 
resembles  16.  In  the  latter,  the  cost  of  the  second  pile  is  8  times 
the  cost  of  the  first ;  in  this  one,  the  surface  to  be  painted  in  th<' 
second  room  is  4  times  that  of  the  first  room,  making  the  cost 
$56.  As  they  may  not  be  familiar  enough  with  similar  surfaces 
to  know  the  ratio,  they  should  find  the  surface  of  each. 

1025.  In  dividing  decimals,  change  the  divisor  to  a  whole 
number.     See  Arithmetic,  Art.  663. 


1026.    2.    XXV  =  25000. 


NOTES   ON   CHAPTER   THIRTEEN  157 

5.  The  furlong  is  seldom  used.  3.7082  mi.  ^  4  -=  .92705  mi., 
the  length  of  one  side.  Multiplying  by  8  to  reduce  to  furlongs, 
we  obtain  7.4164  fur.  Change  the  decimal  part,  .4164  fur.,  to 
rods  by  multiplying  by  40,  obtaining  16.656  rd. ;  .656  rd.  = 
3.608  yd. ;  etc. ;  etc. 

6.  See  Art.  986,  7.  That  selling  price,  $3.54,  must  be  in- 
creased |,  or  $  .59,  to  gain  16|%.     $3.54 +  $.59 -$4.13.  Ans. 

„     qj  5  ^  240  X  38  X  9      ^        1 
7-    ffX — Cancel. 

10.  The  inventors  of  the  expression  "true  discount"  assume 
that  interest  is  not  payable  in  advance.  They  claim  that  a 
borrower  that  promises  to  pay  $380  at  the  end  of  2  yr.  5  mo. 
should  receive  as  a  loan  the  principal  that  will  amount  to  $380 
in  that  time. 

Let      X  =  principal. 

Interest  =  a:  X  —  X —  = —• 
100      12      200 

Amounts  a; +  —  =  380; 
200 

200  a:  +  29  a;  =  229  a;  =  76000  ; 
a;  =331.88-. 
The  principal  (**  present  worth  ")  =  $331.88. 
The  "  true  discount  "  =  $380  -  $331.88  =  $48.12. 

This  "true  discount"  is  the  interest  at  6%  on  $331.88  for 
2  yr.  5  mo. 

The  interest  on  $380  at  6%  for  2  yr.  5  mo.  is  $55.10;  the 
difference  between  the  interest  and  the  "true  discount"  being 
$55.10- $48.12  =  $6.98.  Ans. 

1027.  1.  A  gain  of  25%,  or  J  of  cost,  makes  the  selling  price 
^9,  equal  |-  of  cost;  \  of  cost,  the  present  gain,  is  $9-^5,  or 
$  1.80.     A  gain  of  50%  would  be  $  1.80  X  2  =  $  3.60. 

N.  B.  —  It  is  not  necessary  to  find  the  cost. 


158  MANUAL    FOR   TEACHERS 

2.  $30  in  one  case  represents  f  of  cost,  making  the  gain  $6  ; 
in  the  other  case,  $30  represents  f  of  cost,  making  the  loss  $10. 
Net  loss  $4. 

Note.  —  The  thoughtful  teacher  will  recollect  that  every  member  of  the 
class  does  not  "see  through"  an  example  in  the  same  way,  nor  with  equal 
rapidity.  While  quick  work  should  be  exacted  where  the  question  involves 
but  a  single  arithmetical  operation,  time  should  be  given,  in  problem  work, 
to  pupils  that  do  not  quickly  grasp  the  conditions.  Such  as  can  dispense 
with  unnecessary  figures  should  be  encouraged  to  do  so  as  much  as  possible; 
but  care  should  be  taken  not  to  injure  others  by  requiring  them  to  adopt  a 
short  method  whose  underlying  principles  they  do  not  thoroughl}^  under- 
stand. Each  should,  to  a  certain  extent,  be  allowed  to  use  his  own  mode 
of  "analyzing"  oral  problems  and  of  setting  down  his  written  ones; 
shorter  ways,  however,  being  presented  from  time  to  time  in  the  oral  and 
blackboard  work  of  his  brighter  classmates.  The  scholar  that  reaches  his 
results  by  a  circuitous  course  will,  by  these  models,  be  led  to  see  the  time 
saved  by  shorter  methods,  and  he  will  probably  try  to  master  some  of  them. 

3.  Together  they  have  $300  =  a: +  2 a;. 

6.  $40  in  3|  yr.  =  $12  per  year.  This  is  produced  at  6% 
by  $200.  Ans. 

7.  The  interest  of  x  dollars  for  5  yr.  at  6%  =  •— ;  x  -{-  --,  or 

8.  Yearly  interest  =  $3.  To  obtain  $18  interest  will  re- 
quire 6  yr. 

9.  The  interest  is  $2,  \  of  principal  in  3J  yr. ;  in  1  yr.  it 
is  ^  X  -^  of  principal,  or  -^  =  5%. 

Or,  the  interest  on  $12  @  1%  for  a  year  is  12^,  or  40^  for 
3^  yr. ;  to  obtain  $2.00  interest,  which  is  5  times  as  much,  the 
rate  must  be  5%. 

10.    I  lose  $25,  or  |  of  cost  =  2,^% . 

102a    4.    23i-f(23i-f3})  +  (23i-f-3f  +  3i). 
6.    Let  3a;  represent  the  amount  received  by  one;  and  4a; 
the  amount  received  by  the  other.     Then,  7a:  ^21.63;  a;  =  3.09; 
4a;  =  12.36,  and  3a;  =  9.27.  Ans.  $9.27  and  $12.36. 


NOTES   ON   CHAPTER   THIRTEEN  159 

7.  [(26  X  12)  X  (4  X  12)]  ^  (8  X  4). 

8.  $10-24x2700x890.     ^^^^^^^ 

1500  X  356 

1044.  In  addition  to  the  details  usually  given  in  a  bill,  an 
invoice  shows  the  marks  and  the  numbers  placed  upon  each 
package  shipped.  In  this  invoice,  the  mark  is  given  in  the  first 
column,  and  is  the  same  on  each  case.  The  number  of  each  case 
is  written  in  the  second  column.  Besides  informing  the  receiver 
in  what  case  a  particular  article  is  packed,  it  is  required  by  the 
U.  S.  custom  authorities.  A  certain  percentage  of  cases  in  each 
invoice  is  examined,  and  their  contents  must  agree  in  description, 
quantity,  value,  etc.,  with  the  invoice. 

1500  yd.  @  IJ^.,       £11-14-   4i 
1500   "      "  2 
3000   "      "  IJ   • 
2889   "      "  2^' 


Less  ^V. 


12- 

-10 

23- 

-    8- 

-   9 

29- 

-11- 

-lOi 

£77- 

-  4- 

-llf 

1- 

-18- 

-  n 

£75- 

-   6- 

-   4i 

Duty  @  50% 

Value  in  U.  S.  money,  $366.53.  Duty  @  50%  =  i  value  = 
$183.27,  nearly. 

English  accountants  employ  to  a  great  extent  a  method  by  aliquot  parts 
called  "practice."  The  cost  of  the  fourth 
item  is  obtained  as  here  shown.  The  cost  at  Id 
per  yard,  28Sdd.,  is  easily  reduced  to  pounds, 
etc.  It  is  repeated  to  obtain  the  cost  @  2d. 
Y^d  is  I  of  Id.,  ^^d.  is  \  of  ^^d.,  and  y|  is  ^ 
of  ^^d.  As  the  smallest  denomination  is  the 
farthing,  or  ^  of  a  penny,  |  of  f c?.  is  called  2 
farthings,  or  ^d.,  instead  of  |c?. 

A  similar,  method  may  be  used  in,  reducing  the  above  result  to  U.S. 
money.  See  Arithmetic,  Art.  1005.  The  value  of  £50  is  ascertained  by 
taking  ^  of  $486.65,  or  $  243.325.     £  25  =  ^  of  £50.     5s.  =  £|- ;  etc. 

In  assessing  duties,  the  government  ignores  fractions  of  a  dollar  in  the 
cost  less  than  J;  over  50;*  is  considered  another  dollar.     The  duty  that 


2889  @lcf. 

=  £12-   0-  9 

"  1  " 

12-  0-  9 

"tS" 

4-   0-  3 

"tV" 

1-  0-  Of 

"ft" 

10-  0^ 

£29-11-10^ 

160  MANUAL   FOR   TEACHERS 

would  be  collected  upon  the  foregoing  would  be  50%  of  |367,  or  $183.50. 
Children  should  not,  of  course,  be  burdened  with  such  details ;  their 
answer  should  be  $183.27. 

Some  zealous  teachers  fall  into  the  mistake  of  endeavoring  to  make  their 
pupils  familiar  with  the  methods  of  calculation  peculiar  to  some  callings. 
The  time  assigned  to  the  study  of  arithmetic  can  be  employed  more  profitably 
to  the  scholar  by  giving  him  the  ability  to  handle  ordinary  problems  with 
reasonable  readiness,  than  by  dissipating  his  energies  in  trying  to  make  him 
understand  a  multitude  of  small  matters  that  are  entirely  outside  of  his 
present  experience.  The  average  boy  would  make  a  better  accountant  if 
he  did  not  hear  of  taxes,  partnership,  insurance,  bonds,  stock,  brokerage, 
commissions,  etc.,  during  his  school  life,  provided  the  time  thus  misspent 
were  given  to  elementary  algebra  and  constructive  geometry,  as  well  as  to 
better  work  in  what  are  considered  the  more  elementary  portions  of 
arithmetic. 

1046.    The  ratio  of  3  to  6  is  generally  expressed  as  1  to  2. 

1049.  3.  The  ratio  of  the  price  of  coffee  to  that  of  sugar 
is  ^,  or  5  to  1. 

4.  A  goes  1 J  times  as  fast  as  B.     The  ratio  of  A  to  B  is  J,  or 
5  to  4. 

5.  E's  earnings  are  to  D's  as  4  to  3. 

6.  3  to  4. 

10.    30  to  3  =  10  to  1. 

1050.  6.  One  wheel  makes  70  revolutions  per  second  ;  the 
other  makes  90  per  second. 

7.  The  diameter  =  twice  radius  =  4  ft. 

8.  One  goes  48  mi.  per  hour ;  the  other  goes  the  same  distance 
in  the  same  time. 

9.  Area  of  first  =  6|  X  4  J  ;  of  second  =  4|  X  2^.     Ratio  = 
%^^-     Reduce. 

1051.  2.  Cost  of  farm  =  $75.50  X  156^  X  124.6  ^  160  = 
$9201.52.  Interest  for  1  yr.  on  one-half  of  the  cost  =  $230.04  ; 
$  4600.76  -f  $230.04  =  $4830.80,  amount  of  mortgage.  Deduct- 
ing $500  then  paid,  gives  $4330.80,  balance  due. 


NOTES   ON   CHAPTER   THIRTEEN  161 

4.  Number  of  cubic  feet  of  excavation  =  41|  X  8  X  33  = 
10890.  The  inner  dimensions  of  the  cellar  are  38J  X  30  X  8, 
deducting  from  the  length  and  the  breadth  3  ft.,  which  is  the 
thickness  of  two  walls.  The  number  of  cubic  feet  in  the  cellar 
=  9180,  leaving  10890  cu.  ft.  -  9180  cu.  ft.  =  1710  cu.  ft.  as  the 
contents  of  the  walls.  The  cost  of  the  excavation  is  $^  X  10890 
-^27  =  $201f  ;  to  lay  the  wall,  costs  $15  X  17.1  =  |256.50. 
Total  cost,  $201.66|  + $256.50  =  $458.16J,  or  $458.17.  Ans. 

Note.  —  One  of  the  practices  of  builders  in  some  sections  is  to  take  only 
the  outside  measure  of  walls  in  ascertaining  the  contents.  The  number  of 
cubic  feet  in  this  case,  according  to  their  calculations,  would  be  (41^  +  33 
+  41 J  +  33)  X  8  X  IJ  =  1782  cu.  ft.,  or  72  cu.  ft.  too  many.  By  this  method 
the  four  corners,  measuring  each  1 J  X  1^  X  8,  or  72  cu.  ft.  in  all,  are  in- 
cluded twice.  As  has  already  been  said,  children  should  be  expected  to 
obtain  only  the  correct  results,  leaving  later  experience  to  furnish  informa- 
tion as  to  local  usages. 

5.  For  $107.25  there  can  be  obtained,  at  3J%,  insurance 
amounting  to  $107.25  -^  .0325  -  $3300.  If  this  is  80%  of  the 
value,  the  flour  must  have  cost  him  $3300  ^  |  =  $4125,  which  is 
$4125  -^  500,  per  barrel,  or  $8.25. 

The  algebraic  method  would  be  : 

Let  X  =  the  cost  per  barrel. 

a:  X  500  X  80%  X  3J%  =  107.25  ; 
or,  500a:  X  f  X  ^\\  =  13:c  =  107.25  ; 

X  =  8.25. 

6.  The  bank  discount  is  calculated  on  the  sum  due  at 
maturity,  which  is  $1250+  interest  from  June  12  to  Dec.  15 
(12),  186  da.  (183  da.),  at  5%  =  $1250  +  $32.29  ($31.77)  = 
$  1282.29  ($1281.77).  The  discount  for  30  da.  (including  grace) 
on  $1282.29  at  6%  is  $6.41,  making  the  proceeds  $1275.88. 
The  discount  on  $1281.77  for  27  da.  (no  grace)  at  6%  =  $5.77, 
making  the  proceeds  $1276. 

8.  A  furnished  5  men  for  20  da.  and  6  men  for  15  da., 
which  is  the  same  as  100  men  and  90  men  for  1  da.    B  furnished 


162  MANUAL    FOR   TEACHERS 

the  equivalent  of  120  men  -|-  180  men  for  1  da.  The  money 
received  should  be  divided  on  the  basis  of  190  men  for  A  and 
300  for  B,  490  in  all ;  and  A  should  receive  ^§  of  the  sum,  and 
BfiJofit. 

itt  of  $857.50  =  $332.50,  A's  share; 

itt  of  $857.50  =  $  525.00,  B's  share. 

9.    Arrange  the  work  so  as  to  have  the  required  term  in  the 
last  place : 

A  ditch  (403  X  3  X  3)  cu.  ft.  is  dug  in  (62  X  13)  hr.  by  27  men. 

A  ditch  (750  X  4  X  3)  cu.  ft.  is  dug  in  (250  X  12)  hr.  by  ?  men. 

If  (403  X  3  X  3)  cu.  ft.  are  dug  in  (62  X  13)  hr.  by  27  men, 

1  cu.  ft.  will  be  dug  in  (62  X  13)  hr.  by  ^^^  °^^^"  ^  ; 
^       ^  ^  -^  403  X  3  X  3 

1         r^      -11  k    J       •     1  k     k    27  men  X  62  X  13 

1  cu.  ft.  will  be  dug  in  1  hr.  by -— — - — ; 

^  ^        403x3x3       ' 

etc.     See  Arithmetic,  Arts.  973,  974. 

.  27  men  X  62  X  13  X  750  X  4  X  3     ^ 

Ans.  = Uancel. 

403  X  3  X  3  X  250  X  12 

23a; 
10.    Let  a:  =  price  per  barrel;  — —  =  cost  of  flour.     2J%  of 

23a;         1     .23a;        23a;  •    . 

1  01"  t:::  01  — r-»  —  tt:::  —  commission. 

4  40        4  160 

Then  ^  +  11^=1508.80. 

Note. — The  words  "after  deducting  his  commission"  mislead  some 
pupils,  who  think  that  this  requires  them  to  begin  work  by  deducting  a 
commission  of  ^  of  the  amount  sent.  If  the  purchasing  agent  did  not 
receive  money  in  advance,  he  would  render  his  employer  the  following 
account : 

To  256  bbl.  flour  @  |5.75,        $1472.00 

Commission  at  2^%,  36.80 

$  1508.80 

2|  %  on  the  amount  sent,  $  1508.80,  would  be  2^  %  of  the  cost  of  the  flour, 
1 1472,  and  2^%  of  the  1 36.80  commission. 


NOTES    ON    CHAPTER   THIRTEEN  163 

A  pupil  that  wishes  to  deduct  the  commission  first,  may  work  in  this  way  : 

Let  X  =  commission,  then  40a;  =  cost  of  flour  ; 

41  a;  =  1508.80;  a;  =  36.80. 

$1508.80,  amount  sent,  -|  36.80,  commission,  -|1472  left  for  the  pur- 
chase of  the  flour. 

The  commission  to  be  deducted  is  :^j  of  the  sum  sent. 

1052.  See  Arithmetic,  Arts.  936,  937. 

1053.  Pupils  generally  find  it  most  convenient  to  change  the 
time  to  days,  leaving  the  reduction  to  lowest  terms  for  the  sub- 
sequent cancellation. 

11.    2  yr.  11  mo.  18  da.  =  720  da.  +  330  da.  +  18  da.  =  W/  yr. 

712a;  =3204;  a:  -  4f 
Note.  —  The  teacher  will  notice  that  canceling  100  and  356  by  dividing 
each  by  4,  and  reducing  the  equation  to =  32.04,  is  of  no  advantage. 

The  French  method  of  solving  an  example  like  the  foregoing,  is  to  indi- 
cate all  the  work,  and  then  to  cancel : 

,.  240  X  a;  X  1068  ^  3204  . 

^  ^  100  X  360  100  ' 

(b)  240  X  a;  X  1068  =  ^  X  100  X  360 ; 

3  3 

(c)  ^_?imxX99xm^^_u 

2 

The  first  member  of  equation  (<z)  is  "  cleared  of  fractions  "  by  multiplying 
the  second  number  by  the  compound  denominator,  100  X  360  ;  see  (6).  The 
value  of  X  is  then  found  by  dividing  the  second  member  by  the  compound 
coefiicient  of  x,  240  x  1068 ;  see  (c).  The  result  is  then  obtained  by  can- 
cellation. In  practice,  the  second  equation  (i)  is  not  employed,  the  multi- 
plication by  100  X  360,  and  the  division  by  240  X  1068,  being  indicated  at 
the  same  time. 


164  MANUAL   FOR   TEACHERS 

1054.  See  Art.  948. 

In  states  that  do  not  allow  days  of  grace,  there  will  be  no 
difference  in  these  examples  between  the  words  "  term "  and 
"time."  The  notes  in  19,  21,  and  25  are  assumed  to  be  dis- 
counted on  the  day  they  are  drawn,  which  will  make  the  "term  " 
in  states  allowing  days  of  grace,  3  days  longer  than  the  specified 
time.  Teachers  in  other  states  are  advised  not  to  use  "  days  of 
grace"  in  discount  examples,  no  matter  what  is  the  date  of  the 
note  or  where  it  is  made.  Answers  in  which  days  of  grace  are 
not  included,  are  enclosed  in  parentheses. 

1055.  The  short  methods  previously  studied  should  be  em- 
ployed. For  1,  2,  4,  see  Arithmetic,  Art.  891  ;  6  and  8,  see  Art. 
792;  10,  Art.  717;  12,  Art.  716;  14  and  16,  Art.  791;  17  and 
18,  Art.  958 ;  19,  Art.  758 ;  20,  Art.  910. 

3.    1648x87i  =  |-   of    164800.     See   Arithmetic,    Art.   891. 

5.    2416x875  =  4   of  2416000.     7.    848x125 

on^  J  7    .  1    =i  of  848000.     9.    =  i  of  79200.     13.    =  f  of 
JO^educti  ^^^g^Q      ^^     =3of  imOOO. 

The  pupils  should  be  encouraged  to  employ  these 
methods,  where  practicable.  When  too  many  figures  are  used 
in  blackboard  work,  the  teacher  should  call  the  attention  of  the 
class  to  the  saving  of  time  that  is  rendered  possible  by  the  em- 
ployment of  a  shorter  way. 

1056.  The  use  of  diagrams  will  tend  to  simplify  these  prob- 
lems for  many  pupils.  The  average  scholar  finds  no  difficulty  in 
ascertaining  the  time  difference  when  the  difference  in  longitude 
is  given,  and  vice  versa;  but  the  introduction  of  the  other  ele- 
ments tends  to  confuse  him. 

First  mark  the  meridian  of  0° ;  then  locate  the  two  places, 
writing  under  each  the  longitude,  as  far  as  given  ;  and  above 
each,  its  time,  as  far  as  given.  The  next  step  is  to  find  the  time 
difference  or  longitude  difference,  writing  it  in  the  place  desig- 
nated ;  and  from  it  to  calculate  the  other,  writing  it  in  its  place. 
The  last  step  is  to  calculate  the  required  time  or  longitude. 


NOTES   ON    CHAPTER    THIRTEEN  165 

1.  The  difference  in  time  is  1  hr.  24  min.  The  time  at  A  is 
later;  1  hr.  24  min.  is,  therefore,  deducted  from  1:30  p.m., 
giving  12 :  6  p.m.  as  the  time  at  B. 

2.  The  longitude  difference  is  36°,  which  gives  a  time  differ- 
ence of  36  hr.  -T- 15  =  2  hr.  24  min.,  both  of  which  should  be 
written  on  the  diagram.  As  the  right-hand,  or  more  easterly, 
place  is  later,  the  time  at  B  is  12  M.  -  2  hr.  24  min.  —  9  :  36  a.m. 

3.  A  look  at  the  diagram  shows  the  pupil  which  difference 
must  be  found,  from  which  the  other  is  to  be  calculated.  The  time 
difference  in  this  problem  is  55  min.  30  sec,  making  the  longi- 
tude difference  55'  30"  X  15  =  13°  52'  30".  As  B  has  the  later 
time,  it  is  the  more  easterly.  A  glance  at  the  diagram,  if 
correctly  made,  will  show  the  pupil  that  he  must  deduct  from 
the  longitude  of  A,  156°  48',  the  above  difference,  13°  52'  30",  to 
obtain  the  longitude  of  B,  which  is  142°  55'  30"  west.  Other 
rules  than  those  already  learned  should  not  be  given. 

4.  B  is  52°  36'  east  of  0°,  or  east  longitude. 

5.  The  time  difference  =  101  hr.  ^15-6  hr.  44  min.  The 
time  at  A  =  12  M.  —  6  hr.  44  min.  =  5  :  16  a.m. 

1057.  6.  Long.  diff.  =  9°;  time  diff.  =  36  min.;  9  A.M - 
36  mill.  =  8:24  a.m. 

•    7.    Long.  diff.  =  25°  55' ;   time  diff.  =  1  hr.  43  min.  40  sec. ; 
1  :  45  p.m.  —  1  hr.  43  min.  40  sec.  =  12 :  1  :  20  p.m. 

8.  Time  diff.  =  55  min. ;  long.  diff.  =  13°  45' ;  156°  48'  - 
13°  45'  =  143°  3'  W. 

1058.  Find  the  square  root  of  each  term.  After  extracting 
the  roots  of  13-21,  reduce  the  resulting  improper  fractions  to 
mixed  numbers. 

1059.  6.  Find  the  amount  of  $  1030.05  at  4%  for  2  yr.  9  mo. 
12  da. 

7.    See  Manual,  Art.  1026,  10. 

1060.  See  notes  on  Special  Drills  of  previous  chapters. 


166  MANUAL    FOR   TEACHERS 

1062.  36  X  31  =  (36  x  30)  -f-  36.  36  x  29  =  (36  x  30)  -  36. 
Art.  953. 

1063.  Use  chiefly  as  "  sight "  work.  675  ^  75  =  6f  -^  f  =  27 
^-3;  225^12i  =  2ix8;  150^6^=1^x16;  825-v-37i  = 
8f^|  =  66-^3;  750^62^  =  71^1  =  60-^-5. 

1064.  315xl}4  =  (315x2)-(iVof315);  32  X  39|  =  (32 
X  40)  -  (I  of  32)  ;  etc. 

The  square  of  7|-  (Art.  1032)  =  7  times  7  +  (2  X  7  X  i), 
or  1  time  7  +  ^  =  8  times  7  +  J ;  SJ  X  SJ-  =  9  times  8  +  i. 
The  square  of  75  is  found  by  affixing  25  to  the  product  of  7  by 
8,  5625 ;  85^  =  7225. 

18|  X  5i  =  18|  X  5  (92)  +  i  of  18f  (6j\)  =  98iV  See  Arith- 
metic, Art.  758. 

When  the  divisor  is  a  whole  number,  97^  -f-  3,  etc.,  do  not 
reduce  the  dividend  to  an  improper  fraction  ;  ^  of  97 1  =  32  with 
a  remainder  of  1^,  or  f ;  i  of  f  =  i.  19|^  ^  2|  =  ^  ~^  Y  "= 
96-^12;  etc. 

1065.  These  problems  are  applications  of  the  drills  upon  the 
preceding  page.  Their  solution  involves  simply  the  ability  to 
handle  large  numbers  without  a  pencil,  and  does  not  require  any 
mental  effort  in  determining  the  nature  of  the  operations  required. 
It  is  of  more  advantage  to  the  pupil  to  be  able  to  obtain  the 
results  in  examples  of  this  kind  than  in  the  more  complicated 
ones  usually  given  in  the  higher  grades,  for  which  reason  teachers 
should  be  careful  not  to  omit  them. 

6.  [(30  X  16)  -  (i  of  16)]  oz. 

8.  5700  tenths  h-  19  tenths. 

10.  [(12yr.  x9)  +  (10yr.x6)]-^(9  +  6). 

16.  77^5i  =  77-^-y-  =  '7'7x-j^  =  7x2. 

17.  [(10|x6)4-(ioflOJ)]sq.yd. 

18.  31  doz.  @  15^. 

21.    100  marks  =  $23.80;  50  marks  =  J  of  $23.80. 


NOTES   ON    CHAPTER    THIRTEEN  167 

24.  1^  thousand  X  16  -  20000. 

25.  5000^871  =  56-4--7-=  56  xf  =  8x8. 

1066.  The  value  of  "All  others"  should  be  written  directly 
1  Go  301  4-  63  30  ^^  ^^^  place.  See  Arithmetic,  Art.  384. 
2.  13  143  —  13  14  "^^  obtain  results  with  two  decimal  places 
3        8  576  —       8  58    ^^^^  ^^^^  S^^'^  ^  total  of  exactly  100,  it 

4.  3  826  4-       3  83    ^^^^  ^®  necessary  to  extend  the  division 

5.  3  414  +       3  41    ^'°  three  places,  as  here  shown,  and  to  in- 

6  2  094  —       2  09    c^'^^se  by  1  the  hundredths'  figure  of  the 

7  1113—       111     ^^^^  having   the    largest   figures   in    the 

8  715  —  72    thousandths'    place,    rejecting    the  thou- 

9  3  818  —       3  82    ^andths'  figures  of  the  others.     The  usual 
100  000       1 00  00    ^^^^^^^  0^  calling  5  thousandths  or  over 

1  hundredth,  does  not  always  make  the 
total  correct.  See  8,  in  which  .715—  is  made  .72,  although  the 
thousandths'  figure  is  not  quite  5. 

1067.  7.  (a)  $28,128  X  39^  --  $1104.02.  A  payment  on 
Dec.  21  is  entitled  to  a  rebate  of  11  days'  interest  on  the  fore- 
going, at  7y\%,  or  $2.46,  making  the  amount  actually  paid 
=  $1104.02-$2.46-$1101.56.  If  paid  Jan.  15,  the  interest 
on  $1104.02  at  9%  for  45  da.,  $12.42,  would  be  added,  making 
a  total  of  $1116.44. 

1070.  Permit  the  pupils  to  work  out  in  their  own  way  these 
preliminary  examples.  The  sign  :  :  is  another  form  of  the  sign 
of  equality.  The  rule  for  proportion  is  given  in  Arithmetic, 
Art.  1073. 

1074.  In  oral  problems,  pupils  should  generally  be  permitted 
to  use  their  own  method.  Nothing  will  be  gained  by  requiring 
them  to  use  proportion  in  the  solution  of  these. 

1076.  1.  80%  of  70^.  2.  90%  of  60^.  3.  60%  of  75^. 
8.    90%  of  QQ-lfi. 


168  MANUAL   FOR  TEACHERS 

1077.  11.  The  first  discount  is  the  given  one,  30%  ;  the 
second  is  30%  of  the  remaining  70%,  or  21%  ;  the  total 
single  discount  -  30%  +  21%  =  51%.  12.  20%  +i  of  (100 
-  20)%  =  20%  +  20%  =  40%.  13.  25%  +  i  of  (100  -  25)% 
=  25%  +  15%  =  40%.  The  results  are  the  same  in  12  and  13, 
it  making  no  difference  which  discount  is  taken  first. 

Another  method  of  obtaining  the  single  discount  is  to  ascertain 
the  net  per  cent,  and  to  subtract  this  from  100%.  Thus,  in  11, 
(100  -  30)%  of  (100  -  30)  per  cent  =  70%  of  70  per  cent  =  49 
per  cent.     The  single  discount  =  (100  —  49)%  =  51%. 

107a  21.  30  and  20%  =  (70%  of  80%)  net  =  56%  net. 
40  and  10%  =  (60%  of  90%)  net  =  54%  net.  The  latter,  being 
the  smaller  price,  is  better  for  the  buyer. 

1079.     1.    60%  of  00%  of$250  =  $135.  Ans.     2.   95%  of 

(100  -  x)%  of  800,  or  ^  of  ^^^  ~  ^  of  595  =  684 ;  canceling, 

3800  —  380:^^3^    3800  -  38 ar  =  3420 ;  -38a;=-380;  38a; 

5  '  '  ' 

=  380  ;  X  =  10.  Ans. 


3.  66|%  of  90%  ofo:,  or  I  of  A  of  a;  =  ^=90;  3a;  =  450; 
a;  =  150.  Ans.  $150. 

4.  70%    of  (100-a;)%    of  600  =  378;  -I-x^^^X^ 

lU  lUU  1 

^2100 -21a; ^3^g     2100  -  21  a;  =  1890 ;   -21a;  =  -210; 

5 
21a;  =  210;  a;  -  10.  Ans.  10%. 

Note.  —  Some  pupils  may  prefer  to  use  the  longer 

800     method,  deducting  each  discount  in  turn  to  obtain  the  net 

5  %,  40     price,  and  making  this  equal  to  the  given  net  price.     Thus 

760     in  2  the  operations  would  be  shown  as  here  indicated,  and 

X  0/,.  ^     the  equation  would  be  760  -  ^  =  684,  which,  after  clear- 

,       38  X     ^"8  ^^  fractions,  reduces  to  the  form  given  above,  3800  — 

760 —     38 X  =  3420.     The  first  method  is  shorter  and  less  likely  tc 

give  rise  to  mistakes. 


NOTES    ON    CHAPTER   THIRTEEN  169 

5.  X  -=  70%  of  20%  of  16.  8.    ^  of  y^  of  a:  =  73.50. 

6.  ^of  A  of  a:  =  27.  9.    :r  =  f  of  f  of  200. 

4  f  100  -  a;  ^  5  =  3.20.     10.   -of  ^^^^^ of  1.50  =  .60. 

5  100  2  100 

1084.  2.  There  are  5280  ft.  in  a  mile.  5280  X  176  ^  3520 
=:  number  of  minutes.     Cancel. 

3.  T!KTrA.  =  $9;  1  A.  =$9  ^Tt^f^  =  $9  X  ^f^^  ;  ^A. 
=  $9x^^|-^X^.     Cancel. 

4.  Reduce  both  to  pence. 

5.  A  receives  fffj  of  $576  ;  etc. 

6.  (580  tiles  X  6  X  6)  ^  (4  X  3).     Cancel. 

7.  -^=1500. 
400 

10.  108  (in.)  X  80  X  77  ^  231. 

11.  .625  +  .4375  +  .75  +  .09375  +  2.46  =  312.5  x. 

12.  142.50  ^  (5.25  -  4.95)  =  number  of  tons,  475.  Total 
cost  =  $4.95x475. 

14.  The  distance  traveled  in  1  revolution  is  13  ft.,  the  cir- 
cumference of  the  wheel. 

15.  Three  hundred  forty-nine  thousandths;  three  hundred 
(units)  and  forty-nine  thousandths ;  three  hundred  forty  nine- 
thousandths  ;  three  hundred  (units)  and  forty  nine-thousandths. 

In  dictating  such  numbers,  it  would  be  necessary  to  be  still 
more  explicit. 

16o  4  lb.  6  oz.  12  pwt.  =  1092  pwt. ;  7  lb.  9  oz.  12  pwt.  = 
1872  pwt.  Am.  (£13  8s.  4c?.)  X  1872  -4-  1092  =  (canceling) 
£13  8s.  4cf.  X  y.  If  preferred,  £  13  8s.  4d  may  be  reduced  to 
pence. 

17.  Total  amount  realized  =  $20000  +  $1500  -f  $1000  = 
$22500.  A  furnishes  one-third  of  the  $15000  capital,  and  is 
entitled  to  \  of  $22500,  or  $7500.  Having  drawn  $1500 
already,  he  should  receive  $6000  additional.  B's  share  = 
$  1 5000  -  $  1000  =  $  14000. 


170  MANUAL    FOR   TEACHERS 

18.  The  gain  on  an  article  bought  for  80%  of  its  value  and 
sold  for  120%  of  its  value,  is  40%  of  its  value.  40%  is  ^  of  the 
cost,  80%,  so  that  the  gam  is  50%. 

The  words  "  per  cent '  occur  so  frequently  in  the  foregoing  as 
to  confuse  some  children.  Galling  the  value  100 a;,  the  cost 
is  80  a;  and  the  selling  price  is  120  a:,  a  gain  of  40  a;,  which  is  ^  of 
the  cost,  80a;,  or  50%.- 

19.  $  9  half-yearly,  or  $  18  yearly,  is  the  interest  on  the  differ- 
ence between  $1200  and  $750,  or  $450.     The  rate  is  4%. 

21.  M  does  i  in  a  day;  N,  J;  0,  J;  together,  i  +  i  +  i  in 
1  da.,  or  15  +  12  +  10  _  37  .^  ^  ^^      rp^  ^^  ^^^  ^^^^^  ^^^^     .jj 

60  60 

require  1  da.  -^  f^  =  f^  da.  =  Iff  da. 


The  teacher  should  not  neglect  to  have  her  pupils  make  an 
estimate  of  the  answer  of  each  problem  before  beginning  work. 
They  will  find  in  doing  this,  that  they  can  solve  many  of  the 
problems  without  using  the  pencil.  It  will  be  found  useful  as  an 
occasional  class  exercise  to  require  each  pupil  to  write  on  paper 
at  a  signal,  his  estimate  of,  say,  ten  problems  similar  to  some  of  the 
foregoing.  Anything  that  will  cause  a  pupil  to  read  carefully 
and  understandingly  before  setting  his  pencil  to  paper,  will  by 
of  great  value  to  him  in  problem  work.     See  Art.  981. 

1086.     1.    Let  a;  =  first  cost   of  the   goods.     Then,   ^  = 

122.16  ;  24  a;  =  12216  ;  a;  =  509  ;  cost  =  $509.  The  number  of 
yards  purchased  =  32  X  32  =  J.024 ;  invoice  price  per  yard  = 
$  509  -J- 1024  =  49^1  ^.  Cost  of  the  goods,  including  duties,  etc., 
=  $509  -f  $122.16  +  $40.96  =  $672.12  ;  total  cost  per  yard, 
=  $672.12^1024  =  65^^. 

2.  The  amount  paid  =  $12500  +  9  months*  interest  on 
$3500+18  months'  interest  on  $2600  +  28  months'  interest 
on  $2400. 


NOTES   ON   CHAPTER   THIRTEEN  171 

3.  At  the  end  of  30  da.,  the  provisions  will  last  1200  men 
70  da.  If  half  the  men  are  then  withdrawn,  the  remaining  600 
will  consume  in  30  da.  what  1200  men  would  use  in  15  da., 
leaving  55  days'  provisions  for  1200  men.  If  the  force  is  in- 
creased then  by  900  men,  there  will  be  1500  in  the  garrison. 
55  days'  provisions  for  1200  men  will  last  1500  men  yf^^  of 
55  da.  -=  -I  of  55  da.  =  44  da.  Total  time,  80  da.  +  30  da.  -f 
44  da.  =  104  da. 

4.  The  ad  valorem  duty  =  20%  of  7^  X  267  X  37.  The 
specific  duty  =  5J^  X  267  X  37  X  -||-.  The  sum  of  both  will  be 
the  entire  duty. 

5.  |off  of-2/  =  if;  ifxiofif-f-f 

6.  971%  of  amount  -  $  762742.50. 

9.    The  tea  cost  25^  per  lb.     A  gain  of  15^  =  60%. 

1087.  These  exercises  should  be  omitted  if  not  required  by 
the  course  of  study. 


Face  of  Draft 

+  Premium, 

OR 

-  Discount 

—  Interest 

= 

Cost  of  Draft 

1.          100 

+ 

.02 

= 

X 

2.            X 

X 

= 

499.50 

1000 

3.     1800 

+ 

9x 
5 

18 

= 

1778.85 

4.          X 

+ 

x 
400 

= 

701.75 

5.       200 

— 

.20      - 

1 

= 

X 

6.       600 

+ 

2.25       - 

36a; 

= 

598.95 

7.     1000 

— 

.75      - 

mx 

= 

999.25 

8.     1200 

31a; 
10 

= 

1178.30 

9.       800 

+ 

^x 
5 

3.20 

= 

796.80 

LO.       400 

+ 

.80        - 

X 

10 

= 

400.30 

172  MANUAL    FOR   TEACHERS 


2.    Interest  =  xX  — —  X  -— -  = 


100     360      1000 

3.    The  rate  is  x  dollars  per  $1000  premium;  on  $1800,  the 

premium  = X  a:  =  —     The  interest  on  $  1800  for  60  clays  (al 

^  1000  5  ^  q 

6%--l%  of  $1800  =  $18.     The  equation  becomes  ^+ 1782 

=  1778.85;  —=  1778.85- 1782  =  - 3.15  ;    9:c  =  - 15.75;   x 
5 

=  —  1.75.     The  minus  sign  indicates  that  it  is  a  discount.     Ans. 
$1.75  discount  per  $  1000. 

6.  Let   X  =  time    in   years.      The    equation   becomes   602.25 
-36:r  =  598.95;   -  36  a;  =  598.95  -  602.25  =  - 3.30;   36  a;  = 

3.30 ;  a;  =  II  =  — •     Ans.  J|  year  =  33  days. 
36      360  360"^  ^ 

7.  — .60a;  =^0;  a;  =  0  years;  ^.e.  sight. 

8.  The    interest  =  1200  X-^X  —  =  —-     The   equation 

100     360       10  ^ 

becomes— =21.70;  31a;  =  217;  a;=7.  Ans.1%. 

800  4  a; 

9.  Rate  =  a;    per    M.      Premium  =  ——- X  a;  =  — ;    a;  =  0. 

^«..  Par.  1«*0  ' 


1088.  4.    10  sq^  ch.  =  1  A.  =  160  sq.  rd.     1  sq.  ch.  =  16  sq. 

rd.     1  chain  =  Vl6  rods  =  4  rods  =  16^  ft.  X  4. 

5.  Each  face  contains  1350  sq.  in.  -^-  6  =  225  sq.  in.  Length 
=.V225in.  =  15in. 

6.  Hypotenuse  =  5  in. 

7.  Hypotenuse  =  3 J  in. 

1089.  Another  method  of  subdividing  square  C  is  here 
shown,  Fig.  3.  From  each  side  of  C,  a  right-angled  triangle  is 
cut,  of  the  same  dimensions  as  the  original  triangle  m,  leaving  a 
small  square  X.     In  Fig.  4,  is  shown  a  rearrangement  of  these 


NOTES   ON   CHAPTER   THIRTEEN 


173 


Fig.  1. 


Fig.  2. 


77l\ 

X 

TTV  ^^ 

...       . 

^^v.^  / 

m  /^ 

/  X  7 

Fig.  8. 

B 

A 

Fig.  4. 


Fig.  5. 


triangles,  making  a  polygon  equal  in  surface  to  the  sum  of  the 
squares  A  and  B,  Fig.  5. 

1091.  1.  To  find  -^^,  divide  by  4,  placing  each  quotient 
figure  one  place  to  the  right.  Nothing  should  be  written  beyond 
what  is  shown  in  the  Arithmetic.  To  find  the  interest,  see  Art. 
983,  27. 

2.  1%  each  quarter.  In  dividing  by  100,  the  dividend  is 
repeated,  with  each  figure  two  places  to  the  right. 

3.  3%  each  half  year.  Multiply  by  3,  placing  each  figure 
of  the  product  two  places  to  the  right. 

$1500. 

45. 

$1545.       iyear 
46.35 


$1591.35  1  year. 

1092.  In  some  cities,  the  quotations  of  stocks  give  the  price 
per  share.  A  share  whose  par  value  is  $50  and  which  sells  for 
$48,  is  quoted  in  the  New  York  papers  as  worth  96  ;  i.e.  96% 


174  MANUAL   FOR  TEACHERS 

of  $50.  In  a  few  places,  the  price  is  quoted  as  48,  meaning  $48 
per  share.  In  the  examples  given  in  the  Arithmetic,  the  rate 
always  means  the  per  cent  of  the  par  value. 

1.  (87f  +  i)  %  of  ($  10  X  240). 

2.  Brokerage  ==  \%  of  ($100  X  120)  =  $30.  Value  of  stock 
=  $11460  -  $30  =  $  11430  ;  value  per  share  =  $11430  -^  120 
=  $95.25. 

3.  Let  X  =  par  value  per  share.  Then  (87^  +  \)%  of  (x  X 
150)^5265;  ^  X  150  X  87f  ^o;  X  150  X  351  ^^^^^ 

^  100  100x4 

Usine  the  cancellation  method,  x  =  — — — — — — 

^  150  X  351 

4.  1%  of  $27500. 

5.  102-J%  of  ($25  X  200)  =  selling  price.  Deduct  there- 
from |%  of  ($25x  100). 

6.  His  income  is  18%  of  the  par  value  ($100  X  60)  = 
$1080.     His  investment  is  450%  of  ($  100  X  60)  =  $27000. 

$1080  is  4%  (Ans.)  of  $27000. 

Or,  irrespective  of  number  of  shares,  18%  -:-  4^  —  4%.  Ans. 

7.  The  stock  =  $50  X  4000  =  $200000.  The  rate  of  divi- 
dend =  $ 20000  ^  $200000  -  yV  --=  10%.  Ans.  10%  -^  1.75  = 
5f-%.  Ans. 

8.  $10000  interest  must  be  paid  to  the  bondholders,  leaving 
($47500  -  $  10000)  $37500  to  be  paid  as  dividends  on  $1000000 
of  stock.     37500  -f- 1000000  =  .0375  =  3| % . 

9.  Let  a;  =  brokerage.  ^^'"^^  +  '^  of  25x360  =  15176.25; 
0000  (168i  +  x)  =  1517625 ;  etc. 

10.    Let  X  =  amount  of  brokerage. 

107f  %  of  (100  X  250)  -x  =  26875; 

26937.50  --x  =  26875 ; 

ar  =  62.50;   brokerage  =  $62.50.  Ans.     Etc. 


NOTES   ON   CHAPTER   THIRTEEN  175 

11.  135050  ^1.75i- $20000  =  par  value  of  stock  ;  1\%  of 
$20000 --=$1500.  Ans. 

12.  Mr.  Tower  receives  $30  interest,  and  $100,  the  face  value 
of  the  bond,  in  six  years,  $130  in  all.  His  investment  was  $104, 
on  which  he  has  received  $130  —  $104,  or  $26  interest  in  6  yr., 
or  $4.33J  per  year.     4^  -^  104  =  .04i  =  4i%.  Ans. 

1093.  1.  A  corporation  is  an  association  of  a  number  of 
persons  legally  empowered  to  transact  business  as  a  single  indi- 
vidual. The  charter  specifies  the  name  of  the  corporation,  the 
amount  of  capital,  the  business  it  is  authorized  to  carry  on, 
the  powers  and  privileges  conferred,  etc.  The  stock  is  the  money 
invested  in  the  business  of  the  corporation ;  a  share  is  one  of  the 
equal  parts  into  which  the  stock  is  divided ;  a  shareholder  is  the 
owner  of  one  or  more  shares ;  a  stockbroker  is  a  person  engaged 
in  the  business  of  buying  and  selling  stocks  on  commission  ;  a 
dividend  is  a^ro  rata  division  of  profits  among  the  stockholders ; 
an  assessment  is  a  sum  levied  upon  stockholders  to  meet  some 
unexpected  expenses,  losses,  etc. 

2.  Income  from  bonds  =  6%  of  $125  X  109  ^  1.09  =  6%  of 
$12500  =  $750,  an  increase  of  $  68.75  over  the  previous  income 
of  $681.25. 

3.  Sum  loaned -$59.57-^-1- =  $59.57x6. 

4.  In  3  yr.  at  7%,  $1  will  amount  to  $  1.21 ;  i.e.  a  payment 
of  $1.21  in  3  yr.  is  considered  equal  to  a  cash  payment  of  $1. 
The  "present  worth"  of  the  delayed  payment  =  $4235 -f- 1.21' 
=  $3500,  which  is  a  loss  of  $  3675  -  $  3500  =  $  175.     Ans. 

5.  The  rate  on  one  =  5%  ^  (.98J  +  .OOJ)  =  5  -i-  .981;  on 
the  other  =  6%  -^1.09. 

The  cost  of  a  $100  5%  bond  is  $98.25  + 25 j^  brokerage  = 
$98.50;  the  interest  on  the  bond  is  5%  of  $100  =  $5 ;  the  rate 
is,  therefore,  5%  -f-  .98^  ;  etc. 

6.  The  time  of  the  place  being  1  hr.  later  than  the  time  of  the 
starting-point,  the  traveler  is  15°  east  of  the  latter  place. 


176 


MANUAL    FOR   TEACHERS 


8.  As  the  note  is  stated  to  be  due  in  3  mo.,  which  may  be 
assumed  to  include  days  of  grace  where  allowed,  the  date  of  its 
maturity  is  Sept.  20.  The  amount  of  the  note  for  92  da.,  at  9%, 
^$2455.20.  The  discount  on  this  sum  at  6%  from  Aug.  8  to 
Sept.  20,  43  da.  =  $17.60.  Proceeds  =  $2455.20  -  $17.60  = 
$2437.60.  Am. 

10.   $4800  less  63  days.'  interest,  $50.40,  --  $4749.60. 
($4797.58  less  60  days'  interest,  $47.98,  =$4749.60.) 

1097.     4.    Havmg    found    that    VlO  =  .316  +,    the    pupils 
should  see  that  V.l  =  .316  -f  • 

In  pointing  off,  begin    at   the    units'  place,  pointing  off  two 
places  to  the  right  or  the  left. 
7.    1.6  is  made  1.60.     See  10. 


1098.     7.    Find  the  amount  of  $467.50  at  6%  from  July  5, 
1881,  to  Dec.  19,  1885. 

8.  Amount  stolen  =:  $650.     The   first  sender  is  entitled  to 
^  of  $523.25  ;  the  second,  to  |f  J  of  $523.25  ;  etc. 

9.  Washington,  D.C,  Dec.  1,  1896. 
THE   UNITED  STATES 

To  James  Ryan.  Dr. 


To  Services,  Nov.  19-30,  12  da. 
"  Traveling  Expenses,  12  da.,  $4.00 

$12.50  117.40  $8.25 

"  Stage  Fare     St'mb't  Fare     Telegraph 


1 126  80 


10.  The  account  for  the  first  quarter,  exclusive  of  box  rents, 
is  $124.96.  On  $50,  the  commission  allowed  =$50.  On  the 
remaining  $74.96,  the  commission  is  60%  =$44.98.  Total  = 
$77  +  $50  +  $44.98  =  $  171.98. 

Salary  for  the  year  =  $171.98 +$194.01 +$174.08 +$167.87 
=  $707.94. 


NOTES   ON   CHAPTER   THIRTEEN  17' 


1099.  Base'  +  perpendicular'  =  hypotenuse' ;  11=^ B'  +  P\ 

1.  225  +  64  =  a;' ;  a;'  =  289  ;  x  =  17,  hypotenuse. 

2.  1225  -\-x^  =  1369  ;  x^  =  1369  -  1225  =  144  ;  a;  -  12,  per- 
pendicular. 

3.  x'  +  225  =  1521 ;  x'  =  1521  -  225  =  1296  ;  :^  =  36,  base. 

1100.  11.  Let  X  =  perpendicular.  Area  =  |  (200  +  160)  X 
^  :=  180  a:  -=  32400  ;  x  =  180. 

12.  i(20  +  :r)xl5=225;  §^^±1^  =  225;  300  +  15a;- 
450;  15:^  =  450-300=150;  .r  =  10. 

13.  |(a;  +  a;  +  6)  X  I0=(a;  +  3)X  10-=  10 a; +  30=  150; 
10a;  =  150- 30=  120;  :i;  =  12,  a;  +  6  =  18.  12  rd.  and 
18  rd.  Ans. 

14.  The  base  =  7  rd. ;  the  altitude,  or  perpendicular,  =  24  rd. 
Area  =  1(24  X  7)  sq.  rd. 

15.  Number  of  stones  =  (84  X  36)  ^  (6  X  3).  Cost  =  $li  X 
(28  X  12). 

16.  The  angle  of  90°  indicates  a  right-angled  triangle. 
Hypotenuse  =  20  chains.  Number  of  chains  of  fence  =  12+16 
+  20  =  48.  1  chain  =  66  ft.  =  4  rd.  Number  of  rods  =  48  X 
4  =  192. 

17.  A  perpendicular  let  fall  from  the  upper  right  corner 
would  form  a  right-angled  triangle,  whose  perpendicular  is  40  ft., 
base  30  ft.,  making  the  hypotenuse  =  V1600  +  900  ft.  =  50  ft., 
the  fourth  side. 

The  number  of  yards  of  fence  =  (40  +  70  +  50  +  100)  X  5^-. 
Area  in  acres  =  [^  of  (70  +  100)  X  40]  -4- 160. 

18.  The  diagonal  is  the  hypotenuse  of  a  right-angled  triangle 
whose  other  sides  measure,  respectively,  90  yd.  and  120  yd. 

19.  The  diagonal  =  Vl296  +  729  chains  =  45  chains.  The 
distance  by  the  road  =  27  chains  +  36  chains  =  63  chains. 
The  saving  is  18  chains,  of  22  yd.  each. 


178  MANUAL   FOR  TEACHERS 

20.  A  40-acre  field  contains  6400  sq.  rd.  Each  side  measures 
Vb4U0  rd.  -  80  rd.     The  diagonal  =  V8^~i^W. 

UOl.     1.    The  loss  is  -J-  of  the  cost,  $300. 

2.  If  3  boys  solve  3  problems  in  3  min.,  1  boy  will  solve 
1  problem  in  3  min.,  and  6  boys  will  solve  6  problems  in  3  min. 

Or,  3  boys  will  solve  6  problems  in  6  min.,  therefore  6  boys 
will  solve  6  problems  in  3  min. 

3.  The  3  boys  eat  12  cakes,  4  each ;  for  which  the  third  boy 
pays  12^,  or  3^  apiece.  The  boy  that  brought  7  cakes  supplies 
3  ;  for  which  he  should  receive  9^.  The  other  boy  furnishes  1, 
and  is  entitled  to  3  ^. 

4.  The  cost  of  the  article  is  50^  ;  a  sale  for  $2.00  is  a  gain 
of$1.50,  or300%. 

6.  A  6-ft.  fence  needs  2  posts,  one  at  each  end  ;  a  12-ft. 
fence,  3  posts  ;  a  30-ft.  fence,  6  posts. 

7.  One  half  is  profit,  the  other  half  is  the  cost.  The  profit 
equals  the  cost,  and  is  100%. 

8.  ixX^x==^x''-=60;   3:r'  =  1200;  a;'  =  400;  a;  =  20. 

9.  -1^  =  90. 

400 

10.  The  difference  between  28  and  75,  inclusive,  =(75~  28) 
+  1. 

1102.  2.  The  quantity  of  provisions  for  each  man  =  2^  lb.  X 
20  =  45  lb.  To  last  24  da.,  the  allowance  should  be  45  lb.  h-  24 
=  IJ  lb.  =  1  lb.  14  oz. 

5.  80^  min.  X  112  -j-  46.     Cancel. 

6.  See  Arithmetic,  Arts.  821,  822. 

7.  First  beam  contains  (66  X  10  X  8)  cu.  in.  The  second 
contains  (a:  X  12  X  12)  cu.  in.  The  contents  of  the  latter  =  ^.f;^ 
times  the  contents  of  the  former. 


NOTES   ON   CHAPTER   THIRTEEN  179 

3024  X  66  X  10  X  8 


a:X  12X  12 

=  120. 


924  X  12  X  12 

Ans.  120  in.  =  10  ft. 

9.    The  weight  of  the  provisions  =  3  lb.  x  32  X  45.     Dividing 
by  40  gives  the  daily  allowance  for  the  increased  crew  ;  dividing 
this  by  their  number,  32  -|-  16,  gives  the  allowance  of  each. 
8  lb.  X  32  X  45 
40  X  48 
12.    The   difference  in  deposits  =  $450,  which  sum  produces 
$18  interest.     The  rate  is  4%. 

14.    $7500  at  x%  for  ^  years  produces  $1125  interest. 

''■   ll00°^2J~ll00°f2J='"' 200^200  =  200=^°' 

a;  =8000. 

16.    [tI^  of  (x  +  400)]  -  (^  of  x)  =  SO; 
5  a: +  2000  _  ^  _^  3q 

100  100 

5  a; +  2000 -4  a;  =  3000, 
X  =  1000, 
3:4-400=1400. 
Ans.  $1000  at  4%,  $1400  at  5%. 

VlOO       5/^VlOO      5/ 

19.  A  receives  ff^  of  selling  price ;  he  invested,  therefore, 
ff^of  $600;  etc. 

20.  F  receives  ■^.  E's  share  is  -^  more  than  D's.  If  -^ 
=  $90,  J^  =  $45;    Fsshare, -3^,  =  $315. 

21.  The  bank  receives  at  the  end  of  63  days,  the  sum  loaned, 
$593.70, +  $6.30  interest.  The  problem  is:  At  what  rate  will 
$593.70  produce,  in  63  days,  $6.30  interest? 


180  MANUAL   FOR   TEACHERS 


593.70  x4:rX:^=  6.30. 


100     360 

630  X  100  X  360 
59370  X  63 


6tV¥^- 


22.  The  two  supply  pipes  fill  ^  +  J,  or  f,  of  tank  in  1  hour. 
If  all  the  pipes  are  set  to  work  when  the  tank  is  full,  the  ex- 
haust pipe  takes  off  each  hour  -J-  of  the  tank  more  than  the 
others  supply.  To  empty  the  tank  would,  therefore,  require  6 
hours. 


1103.     7.    The  diagonal  =  Vl37i"' +  137i^ 
9.    Let  100:<;  ==  value.     Buying  price  =  90x;  selling  price  = 
110a:;  gain  =  20 :r,  which  is  |  of  the  cost,  90a;,  or  22|%. 
10.    420^ H-(f^-|^)  =  number. 

1106.  1.  $  500  -J-  j  =  cost  of  one  portion  ;  500  ^  f  =  cost  of 
other  portion. 

4.  If  J  of  farm  is  sold  for  J  cost,  the  selling  price  of  the 
whole  farm  at  the  same  rate  would  be  J  cost  ^  f  =  f  cost,  mak- 
ing the  gain  \  cost  =  12^%. 

7.  10^  per  bu.  of  60  lb.  =  16|^  per  100  lb.,  or  |^  higher 
Ihan  16^  per  100  lb.,  or  -^^  of  16^  higher,  or  4^%. 

8.  See  Art.  1084,  15. 

10.  If  ^  of  the  selling  price  is  profit,  the  cost  must  be  f  of 
the  selling  price  ;  the  latter  is  therefore  -f  of  the  cost.  This  is  a 
profit  of  "I  of  the  cost,  or  66|%.  Ans. 

Or,  a  profit  of  2-fifths  on  a  cost  of  3-fifths  is  f  of  the  cost. 

11.  x-i-^x  =  60i. 


14.    1J%  of  :^=150. 

o 

17.    See  Arithmetic,  Art.  915,  6-8. 

22.  (100  H-  5i)  years.     Tlie  $200  is  unnecessary. 

23.  (96x48x45)^(231x31^).     Cancel. 


NOTES   ON    CHAPTER   THIRTEEN  181 


24.  20  acres  =  3200  sq.  rd.  Each  side  measures  V3200  rd. 
The  diagonal  =  the  square  root  of  the  sum  of  the  squares  of  two 
equal  sides.     The  square  of  each  —  3200. 

The  diagonal  =  V3200  +  3200  rd.  =  V6400  rd.  =  80  rd. 

Note.  —  By  constructing  a  square  on  the  diagonal  of  a  square,  the  pupils 
will  see  that  the  former  will  be  twice  as  large  as  the  latter ;  that  is,  that  a 
square  on  the  diagonal  of  the  above  will  contain  6400  sq.  rd.,  making  each 
side  80  rd. 

In  the  above  example,  the  square  root  of  3200  should  not  be  extracted. 

1107.  6.  Find  the  proceeds  of  $1572.50  for  81  da.  (Omit- 
ting days  of  grace,  the  proceeds  for  78  da.  =  $1552.06.) 

9.    One  costs  85i%  of  $1500;  the  other  costs  102 J  %  of  $1300. 

10.    ($2562.50 -4- 1.025) -^$62.50  =  number  of  acres. 

1.  It  is  frequently  difficult,  for  various  reasons,  to  measure 
the  altitude  of  a  triangular  field.  On  this  account,  a  method  of 
determining  the  area  when  the  lengths  of  the  sides  are  given,  is 
useful,  even  though  the  underlying  principles  be  not  understood. 
A  pupil  can  satisfy  himself  as  to  its  accuracy,  by  calculating  the 
area  of  the  right-angled  triangle  in  3,  and  of  the  isosceles  tri- 
angle in  5. 

2.  The  half  sum  =  |  of  (35  +  84  +  91)  =  105.  The  re- 
mainders are:  (105-35)  70,  (105  -  84)  21,  and  (105  -  91)  14. 
Area  =  Vl05  X  70  X  21  X  14  sq.  ft.  =  1470  sq.  ft. 

3.  This  is  a  right-angled  triangle,  since  2P -f  28^  =  35'^ ;  its 
area,  therefore,  is  J  of  (21  X  28)  sq.  rd.,  or  294  sq.  rd. 

By  the  other  method,  the  area  =  V42  X  21  X  14  X  7  sq.  rd., 
or  294  sq.  rd. 

4.  The  sides  of  one  triangle  measure  39,  52,  and  65  rd.  re- 
spectively. Its  area  =  V78  X  39  X  26  X  13  sq.  rd.  =  1014  sq. 
rd.  The  sides  of  the  other  are  25,  60,  and  65  rd.,  respectively  ; 
and  its  area  =  V75  X  50  X  15  X  10  sq.  rd.  =  750  sq.  rd.  The 
area  of  the  quadrilateral  =  1014  sq.  rd.  -|-  750  sq.  rd.=  1764  sq.  rd. 


182  MANUAL   FOR  TEACHERS 

Each  of  these  triangles  is  right-angled,  AC  being  their  com- 
mon hypotenuse.  Their  areas  are  ^  of  (39  X  52)  sq.  rd.,  and  ^ 
of  (25  X  60)  sq.  rd.,  respectively. 

5.  Since  the  altitude  AQ  Fig.  2,  Arithmetic,  Art.  1263,  of 
an  isosceles  triangle  divides  the  base  into  two  equal  parts,  BC 
=  15  yd.  AC'  =  AB'  -  BC  -  625  -  225  =  400 ;  AC  =20  yd. 
The  area  --  ^  of  600  sq.  rd.  =  300  sq.  rcL 

The  area  by  the  other  method  =  V40  X  10  X  15  X  15  sq.  yd. 

6.  In  the  right-angled  triangle  ACB  (Art.  1263,  Fig.  2),  BC 
=  i  of  96  ft.  =  48  ft ;  ^^  =  64  ft. ;  AC=  V64^  -  48'  ft.  = 
V4096  -  2304  =  1792  ft.     The  area  =  ^  of  (96  x  42.332)  sq.  ft. 

7.  I-  sum  =  9  ft.     Area  =  V9  X  3  X  3  X  3  sq.  ft. 

8.  The  base  =  V70'  -  42=^  ft.  =  56  ft.  Area  =  J  (42  X  56) 
sq.  ft. 

9.  V50^  —  48'  =  14,  the  number  of  feet  in  one-half  the  base. 
See  Fig.  2,  Arithmetic,  Art.  1263. 

10.  The  common  base  of  the  two  triangles  will  be  one  di- 
agonal, 2  in.  The  altitude  of  a  triangle  will  be  one-half  the 
other  diagonal  =  V2'  -  1'  in.  =  V3  in.  =  1.732  in.  The  diago- 
nal =  3.464  in.  The  area  of  each  triangle  =  ^  (2  X  1.732)  sq. 
in.  =  1.732  sq.  in.     The  area  of  the  rhombus  =  3.464  sq.  in. 

11.  The  scholars  should  be  led  to  ascertain  for  themselves  the 
approximate  relation  between  the  diameter  of  a  circle  and  its 
circumference.  Place  a  point  marked  on  the  circumference  of  a 
spool,  or  something  similar,  on  a  given  point  on  the  surface  of 
a  sheet  of  paper.  Roll  the  spool  until  the  point  on  the  circum- 
ference again  touches  the  surface  of  the  paper.  The  distance 
between  the  two  points  on  the  paper  will  be  equal  to  the  cir- 
cumference of  the  circle.  Measure  this  distance  very  carefully, 
also  the  diameter  of  the  circle,  and  ascertain  the  ratio. 

12.  See  Art.  1274,  6-14,  for  the  reason  for  the  rule  for  de- 
termining the  area  of  a  circle. 

13.  The  i  circumference  =  ^  (2a:  X  3.1416)  =  3.1416a: ;  the  J 
diameter  =-  x ;  the  area  =  3.1416a:  X  x  =  3.1416a:'. 


NOTES   ON    CHAPTER   THIRTEEN  183 

14.  3.1416a;' =  area  =  314.16. 

^^  =  100;  :i;  =  10. 

15.  Diameter  =  15.708  ft.  -^  3.1416  -=  5  ft. 

Area  -  [(i  of  15.708)  X  (i  of  5)]  sq.  ft. 

1108.  The  balance,  $851.72,  is  found  by  adding  the  credits, 
and  subtracting  from  $2535.35  in  one  operation.     See  Art.  384. 

1110.     1.    Amount  of  $500,  July  25  to 

April  1,  250  da.,  $520.83 

Amount  of  $100,  Sept.  18  to  April  1, 195  da.,     $103.25 

"  $200,  Feb.  5,    "      "      "  55     "  201.83       305.08 

Due  April  1,  1894,  $215.75 

3.  Amount  of  $600  for  354  da.,  $635.40 

"   $300   "    152   "  $307.60 

"   $200   "      57   "  201.90       509.50 

Due  at  settlement,  $125.90 

5.  In  the  debit  column,  place  the  interest  for  329  da.,  $46.06. 
The  total  of  this  column  is  written  on  the  same  line  as  the  total 
of  the  credit  column,  and  is  $886.06.  The  amount  is  also  written 
as  the  total  of  credits.  The  total  interest  ($28.38)  on  $500  for 
297  da.,  $24.75,  and  on  $200  for  109  da.,  $3.63,  is  written  among 
the  credits ;  and  the  cash  payment  is  ascertained  by  writing  in 
its  place  the  sum  necessary  ($157.68)  to  make  the  total,  $886.06. 
See  Art.  384. 

6.  From  the  amount  of  $725  for  234  da.  +  the  amount  of 
$603  for  174  da.,  take  the  amount  of  $600  for  183  da.  +  the 
amount  of  $300  for  37  da. 

1113.  3.  The  field  contains  160  sq.  rd.  X  7^.  Its  length  is 
(160  X  7i  -^  30)  rd.  =-■  40  rd.  The  diagonal  =  V40''  +  30"'  rd. 
=  50  rd. 

4.  The  rate  per  cent  that  will  produce  $36  interest  in  1-|  yr. 
is  7|.  The  interest  on  $212.50  at  7|%  for  the  given  time  = 
$52.02.  Ans. 


184  MANUAL   FOR   TEACHERS 

Or,  the  problem  may  be  solved  as  follows  (Arithmetic,   Art. 
974): 

Interest  on  $300       for  20    mo.  is  $36, 
"  $212.50   "   40|   "     "  ? 

$36  X  212.50  y  40t_$36  X  212.50  X  204 
300x20  300x20x5 

5.    The  date  not  being  given,  the  number  of  days  is  taken  as 
120  !-  3.     The  proceeds  for  120  da.  =  $  490. 

8.    The  interest  on  $635  for  205  da.  at  5%  --  $18.08. 
Amount  =  $  635  +  $  18.08  =  $  653.08. 


XVII 

NOTES   ON   CHAPTER   FOURTEEN 

1115.  1.  The  total  interest  on  the  given  sums  of  money  is 
equal  to  the  interest  of  $3000  for  1  mo.  As  the  total  sum  is 
11000,  the  interest  of  $3000  for  1  mo.  equals  the  interest  on 
$1000  for  3  mo. 

1116.  3.  Since  the  time  is  required,  the  products  may  be 
expressed  in  years  (or  months  or  days),  and  their  total  divided 
by  the  total  of  the  multipliers. 

2    yr.  X    600  =  1200  yr. 
1^  "    X    500  =    750  " 
1     "    X    300  =    300  '' 
I  <«    X    400  =    300  " 
?  X  1800  =  2550 

Ans.  2550  yr.  ^  1800  =  1  yr.  5  mo. 

5.  [(15  da.  X  200)  +  (30  da.  X  300)  +  (45  da.  X  400)]  ^  (200 
+  300  -f  400). 

6.  [(1  mo.  X  210)  +  (2  mo.  X  210)  +  (3  mo.  X  210)  +  (4  mo. 
X  210)]  ^  840. 

Since  the  sum  due  at  each  period  is  the  same,  the  210  may  be 
omitted ;      (1  mo.  -f-  2  mo.  +  3  mo.  +  4  mo.)  -^  4. 

7.  [(2  mo.  X  320)  +  (4  mo.  X  160)  +  (5  mo.  X  240)  +  (6  mo. 
X  240)]  ^  960. 

Or,  (2  mo.  X  J)  +  (4  mo.  X  i)  +  (5  mo.  X  i)  +  (6  mo.  X  i) 
=  4^^  mo.  Ans. 

8.  (2  mo.  X  tV)  +  (3  mo.  X  i)  +  (4  mo.  X  i)  -f  (12  mo.  X  j\) 
=  7i|  mo.  =  7  mo.  26  da.  Ans. 

185 


186  MANUAL   FOR   TEACHERS 

9.    (0  mo.  X  i)  +  (3  mo.  X  i)  +  (6  mo.  X  i)  +  (9  mo.  X  i)  = 
4^  mo.  Ans. 

10.    [(0  da.  X  300)  +  (30  da.  X  800)  +  (60  da.  x  1000)]  ^  (300 
+  800  +  1000)  -  40  da.     July  1  -f  40  da.  =  Aug.  10.  Ans. 

1117.     11.    The  total  amount  received  -^  number  of  bushels 
sold  =  average  price. 

The   arrangement   of    the   work   may   follow   that   given   in 
Art.  1116,  Problem  3. 

12.  The  first  puts  in  the  equivalent  of  180  cows  for  a  week; 
the  second,  120  cows  for  a  week  ; 

the  third,  180  cows  for  a  week  ;  a  1^x12  =  180 
total  of  480  cows  for  a  week.  The  ^0  X  6  =- 120 
first  should  pay  Jf^  of  $84;  18x10=180 
the    second,   ^f^   of    $84;    the  480  :  180  ::  $84  :  a: 

third,  m  of  $84.  480  :  120  : :  $84  : 2/ 

Proportion   is  commonly  em-  480  :  120  : :  $84  :  z 

ployed  in  working  examples  of 

this  kind.  As  the  whole  number  for  a  week  (480)  is  to  the  first 
man's  number  for  a  week  (180)  so  is  the  whole  rent  ($84)  to  the 
first  man's  share  (x).  The  second  proportion  is  used  to  ascer- 
tain the  second  man's  share  (y) ;  and  the  third,  to  ascertain  the 
third  man's  share  (z). 

13.  A  furnishes  $2000  for  2  yr.  and  $1000  for  1  yr.,  which 
is   the  equivalent   of 


2000  X  2  =  4000 
1000  X  1  --  1000  5000 


$5000  for  a  year.  B 
furnishes  the  equiva- 
lent of  $6000  for  a    3000x2=  6000 


year.      The   total    is  11000  :  5000  ::  $1100  :  A 

$11000  for  a  year,  and  11000  :  6000  ::  $1100  :  B 

the  profits  of  $1100 

are  distributed  in  the  ratios  of  t^VA  ^"^^  tWA'  ^^  indicated  by 
the  proportions  here  given.  A  receive:  4)500  of  the  profits  and 
his  capital  of  $3000,  or  $3500  in  all.     B  receives  $3600. 


NOTES   ON   CHAPTER   FOURTEEN  187 

14.  40a; +  30 (100 -a;)  =  36x100, 
40  a; +3000-  30:^  =  3600, 

40  a; -30  a:  =  3600 -3000, 
10  a;  =  600, 

x  =  QO  =  number  of  bushels  at  40  ^, 
100  —  a;  =  40  =  number  of  bushels  at  30^. 

15.  ^60  X  a;)  +  (50  X  80)  =  52  (a;  +  80). 

60 a; +  4000  =  52 a; +  4160;  etc.; 
that  is,  X  bu.  @  60^  +  80  bu.  @50j^  =  (x  +  80)  bu.  @  52^. 

16.  A  does  ^  in  1  da. ;  B  does  -^^j  in  1  da.  ;  both  do  yu-\--^ 
=  -^Y  in  1  da.  They  finish  the  work,  therefore,  in  12  da.,  and 
receive  $60.  If  A  does  -^  in  1  da.,  in  12  da.  he  does  ^,  and 
should  receive  i|  of  $  60  =  $36.     B  should  receive  $  24. 

17.  C's  capital  of  $4000  for  i  yr.  may  be  considered  as  $2000 
for  a  year.  This,  with  $4000  furnished  by  A  and  B,  makes  the 
capital  $6000.     A  takes  ff^^,  or  ^  of  the  profits  ;  etc.     See  13. 

18.  600  :  180  : :  180  tons  :  A's  share. 
600  :  105  : :  180  tons  :  B's  share. 
600  :  315  : :  180  tons  :  C's  share. 

19.  Let  X  =  number  of  quarts  of  water,  then  40  —  a;  =  num- 
ber of  quarts  of  milk.  40  quarts  of  the  adulterated  article  at  5^ 
per  quart  =  200^.  (40  — a;)  quarts  of  milk  at  6/  per  quart  = 
(240  —  Qx)^.     X  quarts  of  water  cost  nothing. 

200  =  240 -6  a;, 
6a;  =  40, 
a;  =  6|. 
The  can  contains  6-|  qt.  water  and  33^  qt.  milk. 

1120.  1.  $999  =  rentforlyr.  10.  mo.  6da.,or22imo.  The 
rent  for  1  mo.  =  $999  ^  22^  =  $999  X  yfy  ;  for  12  mo.  =  $999 
X  yfr  X  12. 

2.    What  per  cent  of  55  oz.  is  121  oz.  ? 

-^  of  55  =  121;  55a;=12100;  a;  =  220. 
100 


188  MANUAL   FOR   TEACHERS 

3.  The  field  contains  1600  sq.  rd. ;  each  side  measures  40  rd., 
or  220  yd.  ;  etc. 

4.  A  solves  20  per  hour;  B  solves  15  X  f^,  or  16^^,  per 
hour;  both  solve  36^^  per  hour.  To  solve  100  will  require 
(100  ^  36^)  hours. 

6.  Six  faces,  each  containing  (15  X  15)  sq.  in. 

7.  Cost  90jZ^.  See  Arithmetic,  Art.  924,  7.  Selling  price 
=  90^X1.43^. 

8.  March  4,  1861,4-4  yr.  1  mo.  11  da. -- Apr.  15,  1865; 
Apr.  15,  1865,  -  56  yr.  2  mo.  3  da.  =  Feb.  12,  1809.  Am. 

9.  The  selling  price,  $4.50  =  90%  of  cost ;  the  latter  is, 
therefore,  $5  per  barrel.     Loss  per  barrel,  50^;  on  50  bbl.,  $25. 

A  profit  of  6%  =30^  per  barrel ;  the  gain  on  100  bbl.  = 
$30.     Net  gain  =  $5.  Ans. 

10.  60%  of  66|%  =40%.  If  j\%  of  a  number  =  810,  the 
number  =  810  X  -V/  =  2025. 

Note.  —  Observe  the  difference  between  this  example  and  8  of  Art.  1101. 

1121.  1.  Having  used  J  of  ^  box,  the  remainder  —  -J-  of  ^^  box 
=  -jij  box.  If  ^  box  =  $iJ,  a  box  =  $|^  X  14  =  68/2^  X  14  = 
$9.52.  Ans. 

3.  The  40- ft.  ladder  forms  the  hypotenuse  of  two  right- 
angled  triangles ;  C^and  BE,  Arithmetic,  Art.  1250,  Problem  9. 
CA,  one  perpendicular,  measures  21  ft. ;  DB,  the  other,  measures 
33  ft.     AB  is  the  width  of  the  street. 


-4^  =  V40^-2P;  EB  =  VW^-SS'',  AE-\-EB  =  AB. 

4.  16oz.  :12oz.  ::$28:a:. 

5.  A  partnership  is  the  association  of  two  or  more  persons 
for  the  transaction  of  business  on  joint  account.  One  advantage 
of  a  partnership  is  the  employment  of  a  larger  capital,  with  a 
smaller  percentage  of  expenses  than  would  be  the  case  were  each 
member  to  establish  a  separate  business.  The  firm  obtains  the 
combined  business  experience  of  its  several  members,  and  can 


NOTES   ON   CHAPTER    FOURTEEN  189 

utilize  the  services  of  each  in  the  department  in  which  he  can  best 
serve  the  interests  of  the  firm  ;  etc. 

Note.  —  In  the  absence  of  an  agreement  to  the  contrary,  each  partner  is 
entitled  to  an  equal  share  of  the  profits,  and  is  liable  for  an  equal  portion 
of  the  losses;  in  the  examples  given,  however,  the  gains  and  the  losses  are 
distributed  according  to  the  amount  invested  by  each  and  the  length  of 
time  each  one's  capital  remains  in  the  business.     See  also  Art.  977. 

6.  A,  ^\  of  $  13  ;  B,  iV?  ;  C,  ^\  ;  D,  ^^^^  ■  etc. 

7.  The  analysis  of  a  mathematical  problem  or  operation 
should  be  occasionally  used  as  an  exercise  in  composition,  the 
same  attention  being  given  to  penmanship,  spelling,  language, 
and  arrangement  as  in  other  such  exercises. 

1122.     1.    Make  men  the  last  term  ;  Art.  974,  8. 
If  1  be  eaten  in  35  da.  at  24  oz.  daily  by  3600  men, 
2  will  be  eaten  in  45  da.  at  14  oz.  daily  by  ?  men. 

3600  men  X  35  X  24  X  2      ^^no  a 

— _ =  9600  men.  Arts. 

45  X  14 

4.   $3700^(1 -.075). 

8.  See  Arithmetic  Art.  1005,  2.  £500  =  -J  of  $4866.50. 
Take  £250,  £25,  £5,  10s.,  5s.,  2s.  6d,  Is.  3d.,  2d. 

Or,  20)  $4.8665  x  780  = 

12)$  .2433  X    18  = 

$   .0203  X    11  = 

9.  See  Art.  1250,  8.  Calling  the  part  remaining  x,  the  part 
broken  off  will  be  100  —  x.  The  latter  is  the  hypotenuse  of  a 
right-angled  triangle ;  x  is  the  perpendicular ;  40  is  the  base. 

40''-f:r^  =  (100-:r)', 
1600  i-x'  =  10000  -  200  :r  4-  x\ 
200  X  =  10000  -  1600  =  8400, 
a:  =  42, 
100  -  a:  =  58. 

The  length  of  the  part  broken  off  =  58  ft.  Ans. 


190 


MANUAL    FOR   TEACHERS 


By   assuming   x   as   the   length  of  the  part  broken  off,  the 
hypotenuse  —  a;,  and  the  perpendicular  =  100  —  x. 
a;2^40'_^(100-a;^), 
x^  =  1600  +  10000  -  200  rp  +  x', 
200a: -11600, 
X  =  58. 
10.    Number  of  feet  in  length  =  10  ^  (ff  X  H)- 
13.    At  lid.  per  pound,  2240  lb.  are  worth  3360c?.  =  £14. 
Cost  of  120  T.  =  £  1680 ;  the  duty  in  English  money  is   |  of 
£1680  =  £336;  freight  =£30;  etc. 

15. 
Dr.  U.S.  TREASURY   DEPARTMENT.  Or. 


1888. 

1883. 

Jan. 

8 

To  2575  lb.  Twine        .12 

809 



Feb. 

4 

By  Cash 

175 

— 

Apr, 

4 

"    25  doz.  Pens        25.— 

625 

— 

Apr. 

30 

"    Cash 

860 

— 

May 

7 

'•    &45  reams  Paper  2.— 

1290 

— 

July 

15 

"    Cash 

700 

— 

July 

9 

"    46  doz.  Ink            3.— 

135 

— 

Nov. 

5 

"    Cash 

2300 

— 

Oct. 

80 

"    1000  M  Envelopes  2.- 

2000 

— 

Dec. 

81 

"    Breakage 

75 

— 

Dec. 

5 

"   8  doz.  Inkstands  1.97 

15 

76 

Dec. 

81 

"    Shortage 

60 

— 

Dec. 

81 

"   Cartage 

45 

— 

Dec. 

81 

•'   Balance 

759 

76 

4419 

76 

4419 

76 

1884. 

Jan. 

1 

To  Balance 

759 

76 

The  above  represents  the  account  as  it  stands  upon  the  books 
of  Samuel  Adams.  The  debit  column  contains  the  amounts  due 
from  the  Treasury  Department,  and  the  credit  column  contains 
the  sums  received,  etc.  The  account  is  balanced  by  placing  the 
footing  of  the  debit  column  under  each,  and  by  writing  in  the 
credit  side  the  words  "By  Balance"  in  red  ink,  followed  by 
the  sum  necessary  to  make  the  total  of  the  credit  column  equal 
to  the  total  of  the  debit  column.  Red  ink  is  used  to  show  that 
the  money  has  not  been  paid.  The  account  is  reopened  by 
writing  "  To  Balance  "  on  the  debit  side,  followed  by  the  sum 
remaining  due. 

The  statement  rendered  by  Samuel  Adams  to  the  Treasury 
Department  would  be  made  out  as  follows : 


NOTES   ON   CHAPTER    FOURTEEN  191 

Washington,  D.C,  Jan.  2,  1884. 


V.a.  TREASURY   DEPARTMENT, 


In  Account  with  Samuel  Adams. 


1883. 

Dr. 

Jan. 

3 

To  Mdse.,  as  per  bill  rendered 

309 

— 

Apr. 

4 

"     "      " 

625 

— 

May- 

7 

"     "      " 

1290 

— 

July 

9 

u              ..                 ..         ..           .. 

135 

— 

Oct. 

30 

"      "        "     "      "        " 

2000 

— 

Dec. 

5 

..         u           ,. 

15 

76 

Dec. 

31 

"    Allowance  for  Cartage 
Cr. 

45 

— 

4419 

76 

1883. 

Feb. 

4 

By  Cash 

175 

— 

Apr. 

30 

.. 

350 

— 

July 

15 

It        ti 

700 

— 

Nov. 

5 

75.-                  60.- 

2300' 

— 

Dec. 

31 

"    Breakage      Shortage 

Balance  due 

135 

— 

3660 

— 

759 

76 

The  quantities  and  prices  are  omitted,  as  they  have  been  given 
in  the  bills  rendered  at  the  time  the  articles  were  supplied. 

For  the  form  of  this  account  as  it  would  appear  on  the  books 
of  the  Treasury  Department,  see  Art.  1146,  24. 

1123.     1.   5280^15.7. 

The  owner  of  a  cyclometer  should  calculate  the  number  of  revolutions  of 
a  wheel  necessary  to  move  the  index  of  the  cyclometer  over  one  of  its 
smallest  divisions.  The  circumference  of  a  wheel,  26  in.  in  diameter, 
measures  nearly  2^  yd.  Nine  revolutions  of  the  wheel  should  indicate  a 
trifle  over  20  yd.  on  the  cyclometer ;  8  revolutions  should  indicate  a  trifle 
over  y^-j  mile,  17.6  yd. 

3.  The  premium  =  ^  of  f  of  $6000  =  $33.75.  The  loss 
will  be  the  value  of  the  uninsured  one-quarter,  $1500,  and  the 
above  premium. 


192  MANUAL   FOR  TEACHERS 

4.  $1.40x(5ix5^)X(8i^3). 

$1.40x16x11x17       . 
3x2x2x3 
The  rods  are  reduced  to  yards  by  multiplying  by  5J- ;  the  feet, 
by  dividing  by  3. 

5.  Let  X  =  the  smaller  number ;  a;+  1  bu.  2  pk.  5  qt.  =  the 
larger  one. 

x  +  x-\-l  bu.  2  pk.  5  qt.  =  12  bu.  1  pk.  3  qt., 
2a;  =  12  bu.  1  pk.  3  qt.  -  1  bu.  2  pk.  5  qt  =  10  bu.  2  pk.  6  qt., 
x=b  bu.  1  pk.  3  qt  =  the  smaller  number, 
x-\-\  bu.  2  pk.  5  qt.  =  7  bu.  =  the  larger  number. 
8.    Each  side  =  40  rd.     Area  --=  1600  sq.  rd.  =  10  A. 


9.    See  Art.  1097.     V.44'  10. 

10.  7000  gr.  X.17U=^  the  number  of  Troy  grains.     Reduce 
to  pounds,  ounces,  pennyweights,  etc. 

11.  At  what  rate  {x)  will  $324.61  in  2  yr.  7  mo.  13  da.  pro- 
duce ($384.13 -$324.61)  interest. 

324.61  x-^x5||=  59.52; 
100     360 

Canceling  decimals,  30610723  x  =  214272000 ; 

X  =  7  nearly. 

12.  ^:c-140. 

13.  The  field  contains  6400  sq.  rd. ;  each  side  measures  80  rd. ; 
the  diagonal  =  V6400  +  6400  rd. 

14.  See  Arithmetic,  Art.  591. 

15.  One   edge   measures   V256   ft.  =  16   ft.     Solid  contents 
=  (16  X  16  X  16)  cu.  ft.,  or  (256  X  16)  cu.  ft. 

16.  Cost  and  selling  price. 
Cost  and  profit  (or  loss). 
Selling  price  and  profit  (or  lossj. 

17.  92^X21643^60. 


NOTES   ON   CHAPTER    FOURTEEN  193 

18.  i  of  (22|  X  19f )  sq.  ft. 

19.  See  Arithmetic,  Art.  924,  7. 

21.  14  da.  in  Oct.  +  30  + 31  +  31  +  28  + 31  +  24  in  April 
exclusive  of  April  25  =  189  da.     9  tons  will  be  needed. 

22.  112  A.  96  sq.  rd.  =  112.6  A.  Remainder  =  112.6  A. 
-  48.64  A.  =  63.96  A. ;  63.96  =  y^  of  112.6 ;  x  =  6396  -^  112.6 

=  63960  -^  1126  =  56f||.     Arts.  56ff|%. 

23.  By  4-ft.  wood  is  meant  that  the  sticks  are  4  ft.  long. 
This  makes  the  pile  4  ft.  wide.     Cancel. 

25.  See  Art.  1117,  12. 

1  =  18    X    4  -    72  =  1008  fourteenths.  Share  -  if  Jf  -  |^, 

2^12^x    5=    62f  =   880          "  "      =i-io%  =  iih 

3=    6i.xll^    71i  =1001          "  "      =-lUh 

4  =  16    X    9  =  144  =  2016          "  "      =20^6  =  224, 

Total,  349if  =  4905 

26.  -1(48  X  perpendicular)  =  160  rd.  X  13^, 

24  X  perpendicular  =  160  rd.  X  13^, 

,.     1          160  rd.  X27      q^    ;i 
perpendicular  = =  90  rd. 

Hypotenuse  =  V^O"  +  48'  ^d.  =  102  rd.  Length  of  fence 
=  48  rd. +  90rd.  +  102rd. 

1124.  2.  ^  circumference  =  3.1416a;;  ^  diameter  =  a;;  area 
=  3.1416:^1 

Note.  —  The  pupil  should  memorize  the  ratio  between  the  circumference 
and  the  diameter,  3.1416.  After  learning  from  Art.  1274,  6-14,  that  the 
area  of  a  circle  is  equal  to  the  product  of  the  semi-circumference  by  the 
serai-diameter,  and  ascertaining  from  2  that  this  is  equal  to  the  square  of 
the  radius  multiplied  by  3.1416,  the  latter  rule  can  occasionally  be  em- 
ployed.    See  6. 

4.    When   the    circumference   is  a;,  the    diameter  = 


3.1416 
=  079.^8^2 
2     3.1416  X  2      12.5664 


Then  ?  X  — -^^ =  ^ =  .07958:^1 


194  MANUAL   FOR   TEACHERS 

5.  18^X3.1416. 

6.  i2'x  3.1416 -153.9384;  i?'=  153.9384 -^  3.1416  =  49; 
R  =  l.  Am.  7  yd. 

7.  Let  X  =  circumference  ;   then,  from  4,  ———-—■  =  area  = 

12.5664 

198.95 ;     x'  =  198.95  X  12.5664  =  2500.08528;     x  =  50.008. 
Ans.  50  rd. 

8.  The  square  of  the  diagonal,  x^,  —  twice  the  square  of  a 

side.     The  square  of  a  side  is,  therefore,  — ,  which  is  the  area  of 
the  square. 

9.  Let  the  pupil  draw  a  square.  On  its  diagonal,  which 
may  be  called  150  rd.,  draw  another  square.  Produce  two  sides 
of  the  smaller  square  so  as  to  make  diagonals  of  the  larger  one. 
An  examination  of  the  small  square  will  show  that  its  area  is 
one-half  that  of  the  other,  or  ^  of  (150  X  150)  sq.  rd. 

10-12.    See  6,  7,  and  1,  of  Measurements,  Art.  1107. 

13.  See  4.     100  sq.  ft.  ^  12.5664  =  7.958  sq.  ft.  Am, 

Area  —  circumference  '^  X  .07958. 

14.  The  altitude  =  V625  -  49  =  24.    Area  =  (40  X  24)  sq.  rd. 

15.  Find  the  perpendicular,  VlOO^  -  80' ;  etc. 

16.  See  10  of  Measurements,  Art.  1107. 

17.  Calculate  the  altitude.     Area  =  i  (60+  130)  X  altitude. 

18.  See  4,  Measurements.  Art.  1107. 


19.  The  altitude  =  V30"'  -  15'. 

20.  Find  the  area  as  in  1,  Measurements,  Art.  1107.  Divide 
the  area  by  one- half  the  base  to  obtain  the  altitude. 

21.  Area  =  800  sq.  rd. ;  area  of  the  square  constructed  on  its 
diagonal  —  800  sq.  rd.  X  2  =  1600  sq.  rd. ;  length  of  the  diagonal 
=  VieOOrd.  =:40rd.  Am. 

22.  See  Measurements,  Art.  1107,  7,  for  the  area  of  one  of  the 
six  equal  triangles. 


NOTES   ON    CHAPTER    FOURTEEN  195 

23.  (6'  X  3.1416)  sq.  in. 

24.  Its  area  is  one-half  the  area  of  the  square  constructed 
on  the  diameter ;  that  is,  i  of  100  sq.  ft. 

25.  The  area  of  the  sector  is  -^  that  of  the  circle.  Area  of 
circle  =  100x3.1416. 

1125.  2.    See  Supplement  for  the  definitions. 
3.   $6.75is  what  per  cent  of  $2700? 

6.  The  note  is  due  April  13  (10),  1891 ;  the  term  of  discount 
is  102  (99)  da.  1261. 

7.  Do  not  place  the  rate  under  the  principal.    ^%        lo.oo 
See  Arithmetic,  Art.  983,  27.  $276.66 

8.  Make  the  divisor  a  whole  number,  68702050000^48665. 
See  Art.  1007,  7, 

9.  To  yield  $900  per  annum,  the  bonds  must  have  a  face  value 
of  $900^. 045  =$20,000.     Their  cost  will  be  $20,000 x  1.0525. 

10.  See  Art.  1051,  10. 

1126.  10.    734V  mi.  ^2f 

11.  See  Art.  1056. 

12.  $3  X4x4-f-f.  Teachers  should  not  require  pupils  to 
use  pencils  unnecessarily.     See  page  5. 

16.    1  +  68  +  ^  =  .. 

18.  See  Arts.  546  and  1022,  15. 

19.  (7of[(14  +  16+14+16)x8]^li)-f-24.  Cancel.  The 
perimeter  of  the  room  multiplied  by  the  height  gives  the  surface 
of  the  walls ;  -|  of  this  gives  the  number  of  square  feet  remain- 
ing after  the  openings  are  deducted ;  dividing  by  1  J,  gives  the  num- 
ber of  feet  of  paper  needed  ;  dividing  by  24,  which  is  the  number 
of  feet  in  a  roll,  gives  the  number  of  rolls,  or  llj.  As  a  part  of 
a  roll  is  not  obtainable,  12  rolls  must  be  purchased. 


196  MANUAL    FOR   TEACHERS 

1127.  5.    Cellar  contains  (10  X  8  X  2)  cu.  yd. 
6.    (48  X  32)  ^  (16  X -i). 

9.    See  Art.  1100,  19. 

1128.  7.    16  :  20  : :  a: :  25.     See  Arts.  1068-1073. 

13.  ^1^  of  f  of  $18000.     Cancel. 

14.  $120^^. 

15.  Rate,  $8  per  1 1000. 

1129.  See  notes  on  previous  Special  Drills. 

1131.  44  X  22  =  (44  X  20)  +  (44  x  2);  44  X  18  =  (44  x  20) 
-(44  X  2).  26  X  62  =  (26  X  60)  +  (26x2);  26x58  =  (26x 
60)  -  (26  X  2). 

1133.  See  Art.  1064.     49  X  49,  Art.  1032. 

1134.  See  Art.  1065. 
5.    9  times  16  ft. 

12.    175  X  12  hundredths  =  If  X  12. 

14.  143-^-V^  =  -VY_x5-13x5  =  65. 

15.  70%  of  69 -70%  of  70-70%  of  1. 

17.    (29  X  16)  +  26  =  10  more  than  (30  X  16). 

20.  2  X  (87  +  49)  =  [(87  +  50)  -  1]  x  2. 

21.  Each  brick  contains  (i  X  ^  X  i)  cu.  ft.  =  -5^  cu.  ft. 

27.  30  da.  after  April  6  =  May  6 ;  30  da.  thereafter  = 
June  5 ;  30  da.  thereafter  =  July  5 ;  adding  days  of  grace 
=  July  8. 

28.  Each  side  measures  2  yd.  The  surface  of  1  face  =  4  sq.  yd. ; 
of  6  faces  —  24  sq.  yd. 

29.  Without  grace,  the  interest  for  60  da.  is  1%  of  $100  =  $1 ; 
the  proceeds  =  $  99. 

For  3  additional  days,  the  interest  is  ^  of  $1,  or  5^;  the 
proceeds  =  $98.95. 


^■UNITEKSITYJ 

NOTES    ON    CHAPTER    FOURTEEN        """"■"  '1 97 

1135.  See  Art.  1044.  The  mark  on  each  case  is  H.  B.  The 
numbers  of  the  cases  are  5453  and  5454.  The  goods  are  sent 
(consigned)  to  Messrs.  Hamburger  Bros.,  to  be  sold  on  commis- 
sion. 


94i  yd. 

at  2s.  Sd. 

£10 

12s. 

n^ 

140|  yd. 

at  Is.  9d. 

12 

6 

H 

61     yd. 

at  3s.  Od. 

9 

3 

0 

348    yd. 

at  Is.  9c?. 

30 

9 

0 

£62 

10 

Hi 

Less  ^, 

1 

11 

H 

£60     19      8 
The  value  in  U.S.  money  ^  $296.78.     Ad  valorem  duty  at 
50%  =$148.39.     Specific  duty  (295  lb.  +  351   lb.)  =  646  lb. 
@U^  =  $284.24.     The  entire  duty  =  $432.63.  Aiis. 

1137.  Divide  area  by  ^  base.  See  Art.  1107,  Measurements, 
1  and  4. 

1139.  1.  A's  equivalent  =  72,  (6  x  12)  ;  B's  -  70,  [(5  X  11) 
+  (3  X  5)].     Total,  142.     A  pays  J^  of  $175  ;  etc. 

2.  Total  debts  ($  750  +  $  1125  +  $  1245)  -  $ 3120. 

3120:  750::  $1287:  A 
3120:  1125  ::  $1287  :B 
3120:  1245::  $1287:0 

3.  K,  50  X  26  ;  L,  60  X  26  ;    M,  70  x  20  ;  N,  90  X  22  ;  etc. 

4.  Some  persons  prefer  to  employ  smaller  figures  for  the 
capital  invested,  by  dividing  each  by  the  same  number.  A's  can 
be  taken  as  $25;  B's  as  $40  ;  and  C's  as  $50;  or  $5,  $8,  and 
$10  may  be  used. 

5  X  12  =    60 

8x    9=    72 

10  X    5  =  ^ 

182:  60::  $15000:  A's  share. 

182:  72::  $15000:  B's  share. 

182:  50::  $15000:  C's  share. 


198  MANUAL   FOR   TEACHERS 


5.  Counting  each  ox  as 

3  sheep, 

60  X  10  -  600 

50  X  8  =  400 

1000 

75  X  8  =  600 

30  X  7  =  210 

810 

54  X  10  =  540 

10  X  12  =  120 

660 

90  X  12  = 

1080 

3550  :  1000 

3550:  810 

3550:  660 

3550  :  1080 

$  152.50  :  W 
$  152.50  :  X 
$  152.50 :  Y 
$  152.50  :  Z 

6.  220  yd.  were  built  in  11  da.  by  18  men.     480  yd.  will  be 
built  in  18  days  by  ?  men. 

18  men  X  11X480  _  34  men.     6  extra  men.  Am. 
220  X  18 

7.  See  Art.  1122,  1. 

8.  14  men  in   (8J  X  12)  hr.   mow  168  acres.     20   men    in 
(7f  X  11)  hr.  mow  ?  acres. 

Note.  —  A  thoughtless  scholar  will  sometimes  fail  to  see  that  the  15  min. 
should  be  joined  to  8  hr. ;  he  will,  therefore,  compare  8  hr.  with  7  hr.,  and 
15  min.  with  48  min. 

9.  To  do  4  times  the  work  in  ^  of  the  time  will  take  20  times 
12  men,  or  240  men.     (Omit  20  da.) 

10.  At  the  time  they  meet  the  sinking  vessel,  60  men  have 
provisions  for  24  da.  ;  these  will  last  72  persons  20  da. 

72  :  60  : :  24  da.  :  x 

11.  Omit  the  dimensions  of  the  boards.  If  76  are  worth 
$19.76,  50  are  worth  $19.76  X  50^  76. 

1140.  This  table  will  furnish  some  practice  in  addition  and 
division,  and  should  not  be  passed  over.    . 


NOTES  ON  CHAPTER  FOURTEEN 


199 


Fig.  1. 


1141.  The  "developed"  entire  surface  of  a  square  prism  is 
shown  in  Arithmetic,  Art.  818,  20.  In  drawing  the  develop- 
ment of  the  convex  surface,  the  upper  and  lower  squares,  denoting 
the  bases,  will  be  omitted.  The  drawing  should  be  done  witli 
reasonable  care,  to  a  scale  of,  say,  ^  inch  to  the  inch. 

In  making  a  model  of  a  solid,  narrow  strips  for  pasting  should 
be  added,  as  shown  in  Fig.  1.  In  the  development  of  a  triangular 
prism,  the  bases  are  usually  drawn  above 
and  below  the  middle  rectangle,  but 
the  pupils  should  learn  that  they  may 
be  placed  in  other  positions,  one  of  which 
is  here  shown.  It  will  be  noticed  that 
the  pasting  flaps  do  not  form  rectan- 
gles, the  sides  being  inclined  at  an  acute 
angle  to  make  neater  work  in  the  com- 
pleted model. 

The  shape   and   the    arrangement  of 
the   gumming  flaps  for  the  bases  of  a 

cylinder  are  shown  in  Fig.  2.  Interested 
pupils  may  be  safely  left  to  themselves  to 
ascertain  the  length  of  the  rectangle  that  is 
needed  for  the  model  of  a  given  cylinder. 

The  scholars  will  learn  more  geometrical 
facts  while  constructing  these   models   than 
Fig.  2.  they  will  obtain  by  memorizing  many  pages 

of  definitions  or  listening  to  nume'rous  "  explanations." 

6.    The  entire   surface  includes  the  convex  surface  and  the 
surface  of  the  bases. 

8.  If  a:  represents  one  side  of  a  cube,  x^  =  the  surface  of  one 
face,  and  6x^  =  the  entire  surface  =  216  sq.  in.  Ans. 

9.  The  convex  surface  =  4:X^  =  144  sq.  in.  Ans. 
11.    The  perimeter  -  (600 -^  15)  ft. 

Or,  let  X  ==  one  side  of  base  ;  the  perimeter  =  4a: ;  the  convex 
surface  =  4^:  X  15  =  60x  =  600  ;  a;  :=  10;  etc. 


200 


MANUAL   FOR   TEACHERS 


12.  Let  a: -=  the  altitude.  (15 +  15 -(-15+ 15)  X  a;  +  15' +  15* 
=  1650 ;  60  :p  +  225  +  225  ■■=  1650 ;  60  a:  =  1650  -  225  -  225 
=  1200 ;  a:  =  20 ;  the  convex  surface  =  60x  =  1200.  20  in. ; 
1200  sq.  in.  Ans. 

13.  Let  a;  =  one  side  of  base;  4a;  =  the  perimeter;  4a:X  15 
=  60a;  =  the  convex  surface  =  540;  x  =  9.  The  entire  surface 
^  540  sq.  in.  -f  81  sq.  in  +  81  sq.  in. 

14.  Circumference  of  base  =  3.1416  ft.  Convex  surface  = 
(3.1416  X  1)  sq.  ft.  =  3.1416  sq.  ft.  Radius  of  base  = -J  ft; 
area  =  (|  X  -|-  X  3.1416)  sq.  ft,  =  .7854  sq.  ft.  Entire  surface  = 
3.1416  sq.  ft.  +  .7854  sq.  ft.  +  .7854  sq.  ft.  =  4.7124  sq.  ft.  Ans. 

15.  See  Arithmetic,  Art.  1290.  While  pupils  should  be  per- 
mitted to  "  develop "  these 
solids  in  their  own  way,  pro- 
vided it  be  a  correct  one,  they 
should  be  advised  in  making 
drawings  for  models  to  use  a 
pattern  that  will  require  a 
minimum  of  pasting.  While 
Fig.  3  would  serve  for  the  de- 
velopment of  the  entire  sur- 
face, it  would  not  answer  as 
a  pattern  from  which  to  con- 
struct a  hollow  pyramid. 

16.  The  convex  surface  of 
a  pyramid  is  equal  to  the  perimeter  of  the  base  X  -J-  the  slant 
height.  The  slant  height  of  a  regular  pyramid  is  the  altitude  of 
one  of  the  equal  triangles  that  constitute  its  convex  surface. 

18.    One  side  of  base  =  Vl44  in. 


Fio. 


19.  Either  calculate  the  slant  height,  which  is  V2'  —  1*=  VS ; 
or  employ  the  method  given  in  1,  of  Measurements,  Arithmetic, 
Art.  1107. 


NOTES  ON  CHAPTER  FOURTEEN  201 

20.  The  developed  convex  surface  of  a  cone  is  a  sector,  whose 
radius  is  the  slant  height  of  the  cone  and  whose  arc  is  equal  in 
length  to  the  circumference  of  the  base  of  the  cone. 

The  circumference  of  the  base  of  the  given  cone  =  (4  times 
3.1416)  in.  The  circumference  of  the  circle  of  which  the  sector 
forms  a  part,  is  (2  X  6  times  3.1416)  in.,  or  (12  times  3.1416)  in. ; 
the  sector  is,  therefore,  -^  of  the  circle,  and  its  arc  measures  -J  of 
360°,  or  120°.  c^ 

Any  sector  of  120°  will  form  a  hollow  e^i^^^^^^^^X^ 

cone  of  the  proper  proportions.  ^<^  ^^b 

The  base  shown  in  Fig.  4  is  not  required  by  ^^oc:r   ,  ti'>^^y^ 

the   terms  of  this  problem  ;  it  is  merely  intro-  ^^  "vT 

duced  here  to  show  the  development  of  the  en-  U-;^.t«-__j 

tire  surface.     As  it  is  difficult  to  lay  off  the  re-  -^V          J^ 

quired  number  of  inches  for  the  arc  AB,  the  '^^J""!^ 

pupil   will   appreciate  the  foregoing  method  of  ^'^^'-  4. 

determining  the  number  of  degrees  it  should  contain.  The  compasses  or  the 
protractor  may  be  employed  to  construct  an  angle  of  120°  at  C. 

The  convex  surface  of  a  cone  is  equal  to  the  circumference 
of  the  base  X  \  slant  height. 

An  examination  of  Fig.  4  will  show  the  resemblance  between  the 
methods  of  calculating  the  surface  of  a  sector  and  of  a  triangle.  The  area 
of  a  triangle  =  \  (base  x  altitude)  ;  that  of  a  sector  =  \  (base  X  radius). 

22.  The  altitude,  one-half  the  base,  and  the  slant  height,  form 
a  right-angled  triangle ;  and  the  lengths  of  the  two  first  being  12 
in.  and  5  in.,  respectively,  the  length  of  the  latter  is  Vi44-j-  25 
in.,  or  13  in.     The  convex  surface  =  |-  (10  X  3.1416  X  13). 

23.  The  entire  surface  =  [|-  of  (6  X  3.1416  X  10)]  +  (3' X 
3.1416)  =  (30  X  3.1416)  -f  (9  X  3.1416)  =  39  X  3.1416. 

Using  IT  (pi)  instead  of  3.1416,  the  circumference  of  the  base 
=  67r  inches.  The  radius  of  the  sector  representing  the  develop- 
ment, is  10  in.,  and  the  circumference  of  the  whole  circle  =  207r 
inches.  As  the  arc  of  the  sector  must  be  67r  inches,  it  measures 
in  degrees  ^  of  360°,  or  108°. 


202  MANUAL   FOR  TEACHERS 

24.  The  slant  height  will  be  ^  of  6  in.  The  circumference  of 
the  base  will  equal  the  arc  of  the  semicircle,  3ir  inches;  its 
diameter  will,  therefore,  be  3  in. 

1142.  3.  Due  Sept.  (21)  24.  Term  of  discount  from  July 
21  =  65  (62)  da. 

4.  $  600  yearly  interest  represents  a  principal  of  $  10000. 

5.  Length  of  one  fence,  (20  +  20  +  20  +  20)  rd. ;  of  the  other, 
(40  +  10  +  40  +  10)rd. 

6.  The  distance  between  the  center  of  the  first  and  of  the 
last  post  =  10  ft.  X  (11  -  1)  =  100  ft.  Adding  ^  of  the  diameter 
of  each  post,  gives  100  ft.  6  in. ;  and  an  additional  3  in.  at  each 
end  to  fasten  the  wire,  makes  a  total  of  101  ft.  of  wire  required 
for  each  length,  or  303  ft.  in  all.  Am. 

1143.  7.    Troy  weight. 

8.  43|  mo.  (g25)i^  per  mo.  43  quarter  dollars  =^$10.75; 
i  of  25^  =  10^.     Total  $  10.85.  Ans. 

11.  Without  grace,  1%  of  $400,  or  $4.  Ans.  With  grace, 
$4  +  ^of  $4,  or$420.  Ans. 

12.  iof  cost  =  $2;  etc. 

13.  The  principal  is  unimportant.     Time  =  (100  -^  8)  yr. 
17.    (100-^6)  yr.        23.'  i%  of  $1234. 

24.  2%  of  $1234. 

25.  The  number  of  rings  =  60  pwt.  -f-  2^  pwt. 

26.  2°3'xl5.      27.    [Twice  (4  +  3)  X  10]  +  [twice  (4  X  3)]. 
28.  $1.12^  +  iof  $1.12^=-$1.12i  +  $.28f 

30.    Selling  price  =  1  of  cost  =  1^;  cost  =  -J^-^f  =  J^.  Ans. 

1144.  4.  14%  profit  =  7^;  cost  =  7^ -^  .14  =  50^;  selling 
price  —b*lf.  Ans. 


NOTES   ON   CHAPTER    FOURTEEN  203 

17.    Cost  of   350   tons  @  $3.50  =  $1225.     Selling   price  = 

$4.25  X  350  XHU- 

Note. —  The  scholars  should  not  use  pencils  to  obtain  answers  to  problems 
that  can  be  solved  at  sight, 

1145.   See  Art.  1290. 
3.    Area  of  base  =  6  sq.  in. ;  etc. 
5.    See  Art.  1107,  Measurements,  7. 

8.  Changing  given  dimensions  to  inches,  the  number  of  gal- 
lons will  be  36^  X  3.1416  X  66  -^  231. 

9.  1  cu.  ft.  =  if|^  lb.     Cubical   contents  =  (3'  X  3.1416  X 
51)  cu.  ft. 

10,  11.  Careful  pupils  will  be  much  interested  in  ascertaining 
how  closely  their  calculations  as  to  the  contents  of  the  measure, 
agree  with  the  number  of  cubic  inches  it  is  supposed  to  contain. 
There  should  be  231  cu.  in.  -^  8,  in  a  quart.  The  cup  used  must 
be  cylindrical.  Tapering  measures  should  be  left  until  the 
frustum  of  a  cone  has  been  studied,  Art.  1295.  The  paper  box 
used  for  ice  cream,  a  frustum  of  a  pyramid,  can  also  be  employed 
at  that  time.  Some  of  these  measurements  should  be  made  out 
of  school,  and  comparisons  made  as  to  the  results  obtained  by 
different  pupils  and  the  methods  employed  by  them  to  secure 
accuracy.  A  random  measurement  will  not  obtain  the  correct 
diameter  of  a  quart  measure. 

After  calculating  the  altitude  of  an  equilateral  triangle  or  the 
diagonal  of  a  square,  the  pupil  should  draw  the  figure  to  a  scale, 
measure  the  altitude  or  the  diagonal,  and  compare  the  measured 
length  with  the  length  obtained  by  calculation. 

Pupils  should  ascertain  the  weight  of  a  cubic  foot  of  water  by 
weighing  a  quart  of  water,  for  instance,  etc. 

12.    Measure  the  height  to  which  the  water  rises  in  the  box,  etc. 

If  a  solid  heavier  than  water,  is  placed  in  a  rectangular  or  a  cylindrical 
vessel  containing  sufficient  water  to  cover  it,  and  the  difference  in  the  depth 
of  the  water  before  and  after  immersion  is  noted,  the  volume  of  the  solid 


204  MANUAL    FOR   TEACHERS 

can  be  calculated.  It  will  be  equal  to  tbat  of  a  solid  whose  base  is  the  base 
of  the  vessel,  and  whose  altitude  is  the  difference  in  depth  above  mentioned. 

If  the  water  in  a  rectangular  box,  whose  base  measures  5^  by  3  in.,  is 
raised  l^  in.  by  the  introduction  of  a  piece  of  marble,  the  volume  of  the 
latter  =  5|  X  3  x  1|  cu.  in. 

This  method  is  useful  in  determining  the  contents  of  a  solid  of  irregular 
shape. 

13.  The  radius  of  the  base  =  i  of  (25.1328  yd. -f- 3.1416) 
=  4  yd.     Volume  of  cone  =  (4'  X  3.1416  X  i  of  18)  cu.  yd. 

14.  See  Art.  1141,  22. 


15.  The  slant  height  of  the  pyramid  =  V24'^  +  (^  of  14)'.  See 
Art.  1283,  13. 

1146.    6.    [(lOf  ^  X  li)  +  (3f)2^x  If)]  X  10840). 
7.    1000  grams  -^  279  =  weight  in  grams  of  a  10-mark  piece. 
Weight  in  Troy  grains  =  (1000  -^  279)  X  15.432349.     Dividing 
this  result  by  23f\^  gives  the  number  of  U.  S.  gold  dollars. 

1000x15  0  43.2349 
279x23  ©22. 

Note.  —  23.22  is  changed  to  a  whole  number  by  removing  the  decimal 
point  two  places  to  the  right,  and  a  corresponding  change  is  made  in  one  of 
the  numbers  in  the  dividend. 

11.  Interest  on  $237453250  @  3  %=$  7123597.50 

250000000  @  4|%  =    11250000. 
737954700  @  4.  %  =    29518188. 

$1225407950  @x  %=  $47891785.50 
4789178550  -^  1225407950  -  3.9083  + 

The  interest  on  the  entire  amount  at  2^%  would  be 
$30635198.75,  the  saving  being  $17256586.75.  Ans. 

12.  $100  worth  of  stock  costs  $85J.  The  annual  dividend  is 
5%  of  $100,  or  $5.  This  is  (5  -f-  .8575)  per  cent  on  the  money 
invested. 

13.  $8930-h1.11|. 


NOTES  ON  CHAPTER  FOURTEEN 


205 


14.  See  Art.  1122,  15.  The  following  shows  the  account  as 
it  stands  on  the  books  of  the  Interior  Department.  The  items 
that  appear  as  debits  on  Mr.  Well's  books,  here  appear  as  credits, 
and  vice  versa. 


Dr. 

RICHARD 

WELLS. 

Or. 

1S82 

1882 

Jan. 

31 

To  Cash 

885 

— 

Jan. 

1 

By  646  bbl.  Flour  9.45 

6095 

25 

Feb. 

5 

u          tt 

450 

— 

(I 

16 

"  1912  bu.  Oats     .57 

1089 

84 

Apr. 

11 

"      " 

615 

35 

Apr. 

4 

"  92311b.  Bacon    .09 

830 

79 

May 

30 

"      " 

4162 

15 

May 

3 

"   8264  bu.  Corn     .74 

6115 

36 

June 

25 

"    345  lb.  Bacon     .09 

81 

05 

June 

20 

"  825  bbl.  Pork  12.65 

4111 

25 

" 

25 

"    35  bbl.  Vork  12.66 

442 

75 

u 

30 

"  Cartage 

65 

— 

" 

30 

"    Penalty 

75 

— 

it 

80 

"    CashinfuU 

11646 

19 

1830T 

49 

18307 

49 

1148.  Multiply  the  length  in  feet  by  the  width  in  feet  by 
the  thickness  in  inches. 

16.  3f^  (per  ft.)X  15x  16xf  X3. 

17.  The  floor  contains  (36  X  17^)  sq.  ft.,  or  630  sq.  ft.  If 
1-inch  boards  were  used,  630  board  feet  would  be  required.  The 
number  of  feet  of  2^-inch  planks  required  —  630  ft.  X  2^. 

18.  (150  X  13  X  I  X  1)  board  ft.  =  1300  board  ft. 

(60  X  14  X  f  X  2)     "        "  -=  1260  "  " 

(40X15X^X4)     "        "  -1000  "  " 

Total,          3560  "  " 
the  duty  on  which,  at  $1  per  M,  is  $3.56.  Ans. 

19.  The  length  of  the  fence  --  480  ft.  +  360  ft.  +  480  ft.  +  360 
ft.  =  1680  ft.  For  a  fence  4  boards  high,  (1680  X  4)  running 
feet  of  boards  will  be  needed,  or  6720  running  feet.  If  the 
boards  are  ^  ft.  wide  and  1  in.  thick,  the  number  of  board  feet 
=  6720x^x1  =  3360  ft.     Cost  =  $18  X  3.36. 

20.  The  length  of  the  fence  =  (25  -f  100  +  25  +  100)  ft. 
Surface  =  (250  X  6)  sq.  ft.  =  1500  sq.  ft.  As  the  boards  are  1 
in.  thick,  the  number  of  board  feet  =  1500.      Cost  =  $25  X  1.5 


206  MANUAL   FOR  TEACHERS 

==  $37.50.  The  number  of  posts  =  250  -^  64  =  40  ;  cost,  at  25  i 
each,  $10.  The  number  of  running  feet  of  scantling,  two  strips, 
=  250  X  2  =  500  ;  the  number  of  board  feet  =:  500  X  ^  X  2  -  250 ; 
cost,  $18  X  .25  =  $4.50.  Total  cost,  $37.50  +  $10  +  $4.50  - 
$52.  Ans. 

1149.  1.  (a)  A  note  made  payable  to  the  order  of  a  certain 
person  or  to  hearer  is  negotiable  ;  in  the  former  case,  an  endorse- 
ment is  necessary  to  transfer  its  ownership.  A  note  payable  to 
hearer  does  not  require  endorsement.  A  note  payable  to  Charles 
Naumann  (without  the  words,  "  or  order,"  or  the  like)  is  not 
transferable  by  endorsement.  If  Charles  Naumann  wishes  to 
sell  the  note,  he  must  assign  his  interest  in  it  by  another 
document. 

Note.  —  The  above  is  the  general  rule;  in  some  states  there  are  special 
laws  bearing  on  the  subject: 

In  Alabama  and  Kentucky,  a  note  to  be  negotiable  must  be  payable  at  a 
fixed  place;  in  Indiana  and  Virginia,  at  a  bank;  in  West  Virginia,  at  a 
bank  or  public  office.  In  Pennsylvania,  it  should  contain  the  words 
"  without  defalcation  "  ;  in  New  Jersey,  "  without  defalcation  or  discount"; 
in  Missouri,  "negotiable  and  payable  without  defalcation  or  discount." 

(h)  A  person  unable  to  write  hfs  name,  makes  his  mark,  as 
shown  below,  in  the  presence  of  a  witness  : 

his 

Witness:  William  X  Devers. 

Theodore  H.  Ficklin.  '"="'' 

3.  In  old  deeds,  the  contents  of  a  farm  are  given  in  acres 
(A.),  roods  (R.),  and  poles  (P.),  the  rood  being  \  acre,  and 
containing  40  poles,  or  square  rods.  In  long  measure,  the  word 
pole  is  occasionally  employed  instead  of  rod. 

6.  2a:-a:  +  |  +  |+18;  etc. 

7.  The  distance  between  the  ships  is  the  hypotenuse  of  a 
right-angled  triangle,  whose  other  sides  are  72  mi.  and  128  mi., 
respectively. 


NOTES   ON    CHAPTER    FOURTEEN 


207 


8.  The  first  capital  =  1 3500 
orf  of  $2500  =  $1500;  etc. 

9.  See  Art.  1026,  10. 
10.    See  Supplement. 


1.40 -$2500;  0  put  in  ||o 


100. 


50,  the  area  of 


1150.     3.    :r' +  a;' =  hypotenuse^ 
the  inscribed  square.  Ans. 

4.  Area  of  circle  =  (5  X  5  X  3.1416)  sq.  in.  =  78.54  sq.  in. 

5.  Arc  of  90°  --  i  (10  X  3.1416)  in.  =  7.854  in.  ;  the  chord 
=  V50in. 

6.  Arc  of  90°  in  a  circle  whose  radius  is  10  in.  =  15.708  in. 
Area  of  sector  =  |  of  (15.708  X  10)  sq.  in.  =  78.54  sq.  in.  Ans. 
Area  of  triangle  = -|  of  (10  X  10)  sq.  in.  =  50  sq.  in. ;  area  of 
segment  =  78.54  sq.  in.  —  50  sq.  in.  =  28.54  sq.  in.  Ans. 

8.  i^'x  3.1416.  Ans. 

9.  Area  of  outer  circle  in  square  yards  =  (15'  X  3.1416)  ^  9  ; 
of  inner  circle  =  (10'  X  3.1416)  -f-  9. 

10.  (125  X  3.1416)  sq.  ft. 

11.  [(6'~3')x3.1416]sq.  in. 
is  6  in. ;  of  the  inner  circle,  3  in. 

12.  36  X  3.1416  :  9  X  3.1416 
=  4:1. 

13.  [(30  X  30)  -  (20  X  20)] 
sq.  ft. 

14.  Dividing  the  walk  into 
four  trapezoids,  as  in  Fig.  5,  the 
area  of  each  will  be  [\  of  (30 
+  20)  X  5]  sq.  ft.  =  125  sq.  ft. 
The  broken  line,  XY,  drawn  mid- 
way between  the  parallel  sides 
measures  25  ft. ;  the  whole  length 
of  the  center  lines  is  100  ft.  The  area  of  the  walk  is  (100  X  5) 
sq.  ft.    • 


The  radius  of  the  outer  circle 
so 


\x 

Y^ 

l\ 

/I 

is 

to 

30 

ai 

V- 

\l 

A          «-          x\i 

30 

Fig.  .5. 


208  MANUAL   FOR   TEACHERS      • 

The  area  of  the  circular  frame  in  11  can  be  ascertained  in  the  same  way. 
The  center  of  the  frame  is  4J  in.  from  the  center  of  the  glass.  The  mid- 
dle line  of  the  frame  =  (3.1416  x  4^  x  2)  in.  =  28.2744  in.  The  area 
=  (28.2744  X  3)  sq.  in. 

15.    The  area  of  the  first  =  240  sq.  ft. ;  of  the  second,  960  sq.  ft. 

U53.  18.  The  surface  of  the  sphere  =  (4  TrXi)  sq.ft.  =  3.1416 
sq.  ft. ;  the  convex  surface  of  the  cylinder  =  3.1416  sq.  ft. 

19.  The  entire  surface  =-  3.1416  sq.  ft.  +  (2  X  J  X  3.1416)  sq.  ft. 
=  1|  times  3.1416  sq.  ft. 

20.  4   times  (^   circumference)  10  X  (^  diameter) 


10 


400  .       .  .     , 

surface  in  square  inches. 


3.1416 


3.1416 


1154.     1.    Rate  per   cent  =  18750  -^  12500  =  If      1^%   of 

$6000  =  $90. 

3.  The  price  of  silver  is  now  given  by  the  ounce. 

4.  The  interest  on  $200  at  4^%  =  $9.  $9  ^  .08  = 
$112.50.  Ans. 

5.  $2100  +  $4400  +  $13000  +  $7200  (90%  of  $8000)  = 
$26700  =  total  assets.  Total  liabilities  =  $1625  +  $5625  = 
$7250.  $26700  -  $7250  =  $19450.  To  this,  add  the  amounts 
withdrawn,  $850  +  $1075=  $1925,  making  the  sum  of  the 
capital  and  profits  $21375.  Of  this,  H  is  entitled  to  i,  $7125, 
less  the  amount  withdrawn  by  him,  $1075,  or  $6050. 

1157.     9-11.    Find  the  cube  root  of  the  numerator  and  of  the 

denominator  separately. 

12-15.    Reduce  to  an  improper  fraction  before  extracting  the 
cube  root ;  then  reduce  the  root  to  a  mixed  number. 

1159.     2.    [f  X  3.1416  X  (f  X  f  X  f)]  cu.  in.     Cancel. 

3.  The  volume  of  the  first  in  cubic  inches  =  ^  tt  X -j- =  |- tf. 
The  volume  of  the  second  =  ^ir  xl=  ^ir. 

4.  The  volume  of  the  sphere  =  .5236  cu.  in. ;  the  volume  of 
the  cube  =  1  cu.  in. 


NOTES   ON    CHArTER    FOURTEEN  209 

6.  The  ball  contains  .5236  cu.  ft.  Its  weight  =  (.5236  X  7.5 
X  1000  -^  16)  lb. 

1160.  1.  40  tons  were  to  be  moved;  there  remain  22  tons 
to  be  moved  in  3  da.  In  6  da.  18  men  moved  18  tons,  so  it 
requires  22  men  to  remove  22  tons  in  6  da.,  or  44  men  to  remove 
22  tons  in  3  da.     44  men.  Ans. 


3.  A  diagonal  on  the  floor  measures  V40^ -f  30^  ft.  =  50  ft. 
A  diagonal  on  one  wall  measures  a/40^~+T4^  ft.  =  42.38  —  ft. ; 
on  the  other  wall  it  measures  V3(F+T¥^  ft.  ==  33.11-  ft. 

4.  Selling  price,  |6500  +  15%  of  $6500  =  $6500  +  $975 
=  $7475. 

As  no  date  is  given,  the  note  may  be  assumed  to  be  for  120  da., 
or  123  da.  including  grace.  Proceeds  of  $7475  for  123  da. 
=  $7321.76.  Profits  =  $7321.76  -  $6500  =  $821.76.  Ans. 
Without  grace,  the  proceeds  would  be  $3.74  more,  making  the 
profits  $825.50.  Ans. 

5.  Let  X  =  face  ;    =  premium  ;  bank  discount  =  x  X  — — 

8        11a;  400      ^  360 

X = (includina;  grace). 

100      1500^  ^^        ^ 

13^  _  11^^4265, 
^  400      1500 

6000  x-^195x-4:4:x  =  25590000, 

6151 X =  25590000, 

a;  =  4160.30-.     $4160.30.  Ans. 

Without  grace,    x  +  ^-~-:^  =  4265  ;  etc. 
4UL)        ioU 

7.  Longitude  difference  =  48°  24';  time  diflference  =  48  hr. 
24  min.  ^  15  =  3  hr.  13  min.  36  sec.  As  the  more  easterly- 
place  has  the  later  time,  the  watch  is  fast.     Ans. 

The  above  result  is  based  upon  the  assumption  that  the  "  sun  time  "  of 
each  place  is  used.  The  difference  in  the  "standard  "  time  of  the  two  cities 
is  3  hr. 


210  MANUAL   FOR   TEACHERS 

9.    Jof|of|ofa;  =  6;  ^  =  6;  a;=192. 

10.  ^-f-|^a;-|+1040; 

40a;  +  8a;  ==40 a:  -  5  a:  +  41600 ; 

13  a;  =  41600;  a;  =  3200.  Ans.  $3200. 

11.  2050:  a;::  41:  69.     Cancel. 

14.  $  1450  -  i%  of  $  1450  -  63  (60)  days'  interest  on  $  1450 
at  5%  =  cost  of  draft. 

17.  A  can  do  ^  in  1  da.  ;  B  can  do  ^;  both  can  do  -}^  in 
1  da.,  and  can  do  the  whole  work  in  T^-  da.  Ans. 

19.    See  Art.  1150,  4. 

1161.  The  definitions  and  principles  called  for  throughout 
this  chapter  should  be  formulated,  as  far  as  possible,  by  the 
pupils,  the  latter  being  led  through  the  teacher's  questioning  to 
see  their  mistakes,  and  to  make  the  necessary  corrections.  If  this 
preliminary  work  is  done  as  it  should  be,  the  scholars  will  be 
ready  to  understand  the  definition  finally  given  by  the  teacher. 
Too  much  time,  however,  should  not  be  wasted  on  formal  defini- 
tions, as  they  are  of  next  to  no  help  to  a  pupil  in  his  mathe- 
matical work,  and  it  is  very  unlikely  that  he  will  ever  be  called 
upon  to  use  them  in  after  life.     See  Supplement. 

1.  (c?)  A  decimal  fraction  is  frequently  defined  as  one  whose 
denominator  is  10  or  some  power  of  10.  In  this  place,  however, 
the  expression  is  used  as  synonymous  with  "  decimal."  The 
rule  asked  for,  refers  to  the  method  of  "pointing  oflf"  the 
product. 

7.  If  all  three  are  opened,  they  will  fill  (yV  +  i^i)  ^^  iV 
of  the  cistern  in  1  hr.  To  fill  the  whole  cistern  will  require 
15  hr.  Ans. 

9.  As  they  meet  in  15  hr.,  they  must  approach  each  other  at 
the  rate  of  (105  -4-  15)  miles  per  hour,  or  7  mi.  As  one  goes  3  mi. 
per  hour,  the  other  must  travel  7  mi.  —  3  mi.  =  4  mi.  per  hour. 


NOTES   ON   CHAPTER    FOUKTEEN  211 

18.  A's  capital  at  end  of  year  is  $8000,  B's  is  $10000,  O's, 
3000:  total,  $21000.  Profits  -  $18000  +  $12000  -  $21000 
-$9000. 

A,  10000  X  i  (yr.)  =  5000 

8000x1  -4000    9000 

B,  6000  X  i  =  3000 
10000  X -I-          =5000    8000 

0,  8000 


20000  :  9000 
20000  :  9000 
20000  :  9000 


9000 :  4050 
8000  :  3600 
3000  :  1350 


A  is  entitled  to  his  capital  at  the  end  of  the  year,  $8000,  and 
$4050  profits,  or  $  12050.  As  he  receives  $  12000  worth  of  goods, 
his  cash  receipts  are  $  50.  B  receives  $10000  +  $  3600  =  $  13600. 
C  receives  $3000  +  $  1350  =  $4350. 

19.  f  +  |of  ^  +  20  +  ^  =  :r; 

f  +  f  +  ^  +  20  =  a:;  etc. 

20.  He  sold  800  bbl.  at  $7.50,  which  amounted  to  $6000. 
The  cost  was  $7200  +  $312  +  $350  -=  $7862.  His  profits  be- 
ing $138,  he  must  have  received  $7862  +  $138  =  $8000.  As 
the  flour  realized  only  $6000,  he  must  have  received  $2000  from 
the  railroad  company. 

28.  16%  of  the  person's  money  =  20%  of  $  160000  ;  etc. 

29.  4  men  do  -|  of  the  work  in  60  hr. ;  to  do  the  remainder, 
they  would  need  120  hr.  ;  and  1  man  would  require  480  hr. 
There  are  8  da.  of  10  hr.,  or  80  hr.,  in  which  to  finish  it ;  6  men, 
therefore,  will  be  needed  to  complete  it,  or  2  men  additional. 

30.  Cost  =  $100 +  $5 +  $50 +  $20  =  $175.  Selling  price, 
$  300  less  10  %  ($  270)  -  $  20  =  $  250.     Profit,  $  75. 


212  MANUAL   FOR  TEACHERS 

If  the  commission  merchant  had  refunded  $  20  before  making 
returns,  his  commission  would  have  been  10%  of  $280,  or  $28; 
and  the  gain  would  have  been  $  77. 

36.    {h)   See  Arithmetic,  Art.  668. 

38.  A,   5000  X  i-  =  2500 

3000  x\^  1500   4000 

B,  6000 

0,   4000  X  i  =  2000 

12000  X  I  =  6000   8000 

18  :  4  ::$6000  :  A,  etc. 

39.  Proceeds  of  $12000  for  93  da.  =  $11814.  Amount  of 
$  10000  for  6  mo.  at  6%  =  $10300.  Sum  remaining  -  $  11814 
-$10300  =  $1514. 

Without  grace,  the  proceeds  of  90-days  note  =  $11820;  sum 
remaining  =  $  1520. 

40.  Cost,  $40  each.  10  were  sold  @  $  44  each  =  $  440 ;  10 
@  $  46  each  =  $  460  ;  15  were  sold  for  $  100  :  total  $  1000.  To 
obtain  $900  for  remaining  15,  he  must  charge  $60  each. 

1164.  1.  (30  +  30  +  31  +  30  +  31  +  31  +  30  +  31  +  30+17) 
da.  =  291  da.     Exact  interest  =  $  400  X  ^  X  |f f 

0  da.  X    630  =  0  da.         4.    Business  men  generally  find 

243   "    X    820  =  199260   "      the   number    of    days'   credit   to 
274   "    X    950  =  260300   "      which  each  item  is  entitled.    $820 

2400    )459560  da.    ^^  ^"^  ^^^^^  ^'  °^  ^^^  ^^'  ^^*^^ 
— ,g,   ,    .        July  5;  $920  is  due  April  5,  or 

^'     274  da.  after  July  5.     The  equated 

time  is  191  da.  after  July  5,  or  Jan.  12.  Ans. 

Using  months  instead  of  days,  the  average  term  of  credit  is 

found  to  be  6  mo.  9  da.,  nearly,  making  the  equated  time  Jan.  14. 

6.  7500  :  A  : :  1200  :  250 

7500  :  B  : :  1200  :  950. 


NOTES  ON  CHAPTER  FOURTEEN  213 

6.  Bank  stock  pays  (25  X  7|-  ^  85)  per  cent  interest  semi- 
annually ;  the  railroad  stock  pays  (25  X  3  ^  31)  per  cent  interest 
semi-annually. 

9.  The  bases  are  equilateral  triangles,  each  side  of  which 
measures  5  in.     Art.  1107,  7,  Measurements. 

1165.     1.    See  Supplement. 

Quantity  is  anything  that  can  be  measured. 

See  Arithmetic,  Art.  1072. 

2.  |of  iiX^XyVxfxAxixV--     Cancel. 

3.  Time  difference  =  2  hr.  37  min.  33  sec. 

11  A.M.  1  :  37  :  33  P.M. 


W. 


?  83°  3'  0° 

Longitude  difference  =  ? 

The  longitude  difference  =  2°  37'  33"  X  15  -  39"  23'  i5". 
Longitude  of  San  Francisco  =  83°  3'  +  39°  23'  15"  =  122°  26' 
15"  west. 

5.  a;  +  2a:  +  5a:  =11480. 

6.  The  equated  time  for  the  payment  of  $600  is  [(200  X  1) 
+  (400  X  2)]  yr.  -^  600  =  1|  yr.  The  present  worth  of  $600 
due  in  If  yr.  =$600^  1.1  =  $545.45+. 

Another  way  is  to  calculate  the  present  worth  of  each,  and  to 
add  the  results  :  ($200  ^  1.06)  +  ($400  ^  1.12)  =  $  188.68  + 
$357.14  =  $545.82.  Ans. 

The  latter  is  the  more  consistent  way  inasmuch  as  it  employs 
the  "  present  worth  "  method  throughout.  The  first  solution  uses 
the  "  present  worth  "  method  to  calculate  the  value  at  date,  oi 
$600  whose  equated  time  has  been  found  by  the  "interest" 
(bank  discount)  method. 

7.  Selling  price  =  $120 +15%  of  $120  =  $138.  Asking 
price  =  $138 +$12  =  $150.     I  threw  off  $12  from  $150,  or  8%. 

8.  Arithmetic,  Art.  1250,  8.  ^i)  =  39,  AC  =  62,  BO 
=  V52^  +  39^  =  65  =  £0.  Height  of  tree  =  52  ft.  +  65  ft. 
=  117  ft. 


214  MANUAL   FOR  TEACHERS 

9.    \/l0.125000  =  ^ns. 

10.    If  I    gain  =  j\  cost,  the  gain  =  ^^  cost  X  f  =  J  cost  = 
16|  per  cent. 

1166.  4.    [1^(^  +  A)]weeks. 

15.    B's  gain  of  $1400  is  -^  of  total  gain  ;  ^  of  total  =  $200; 
A's  gain,  ^  of  total  =  $  1000. 

1167.  4.    Let  X  =  cost  per  barrel. 

75%  of  500  a;  X. 02^  =  80.85. 

5.  A  does  ^  in  1  da. ;  B  does  ^^  in  1  da.  A  does  -Jf  of  the 
work,  or  |,  and  B  does  -f^  of  it,  or  f ,  leaving  -|  to  be  done  by  C 
in  4  da.  To  do  the  whole  work  C  would  require  4  da.  ^  | 
-  18  da. 

6.  A  ditch  20  yd.  X  18  in.  X  4  ft.  is  dug  in  (3  X  10)  hours  by 

72  men.     A  ditch  30  yd.  X  27  in.  X  5  ft.  is  dug  in  (9  X  16)  hours 

by  ?  men. 

72  men  X  3  x  10  x  30  X  27  X  5 


20  X  18  X  4  X  9  X  15 


Ans. 


7.  £2400  income  is  produced  at  3%  by  bonds  whose  face 
value  is  £80000.  Their  cost  =-- £80000  X  .94f  =  £75500 - 
$367433J,  12%  of  which  =  $44092.00. 

$4.86|  X  (2400  ^  .03)  x  94|  x  .12. 

8.  a;  -  —  =  30  +  30%  of  30-39. 

100  ^ 

100  :c- 40  a;  =  3900;  etc. 

10.  [(15  4-  10  +  15  +  10)  X  9f]  +  (15  X  10)  =  number  of 
square  feet  in  the  walls  and  ceiling  =  637^  sq.  ft.  =  10^  sq.  yd. 
The  cost  =  21c?.  X  ^^  --  1487|c?. ;  etc. 

116a  1.  J  +  i  =1^-1.41666  +  ;  ^  +  ^+^  =  1^=^ 
1.41429  ~. 


NOTES  ON  CHAPTER  FOURTEEN  215 


.7409375  -i-  237100  = 

.007409375 

-^  2371. 

See 

Arithmetic, 

rt.  668. 

2      ^i 

,  2     fb 
^^3^U 

■H 

4'^ 

13^3" 

10 
~13' 

2 
"3^ 

-2A  = 

2 
3"^ 

25_2 
12~3^ 

12 
25 

8 
25' 

8i-7f^ 

=-v- 

^¥- 

41- 

-39^|J 

10      8 

41. 

_  750 +  312 

-1025 

37 

13  "^25 

'39' 

975 

975 

975 

_2. 
3' 

37 
975a: 

_2 
3' 

Clearing 

;  of  fractions,  37  ^  650 

^;  ^=-/3V- 

Ans. 

\i  of  ^r  of  I  of  tJ^  X  if  X  H  X  -V- 
a:  =  smaller  number ;  ^  +  t^^  =  larger. 
a;  +  :^  +  _7^^ij|;    126a:+126a:  +  49=:113;    252a:=113- 

49^.64;   a:=2V2-if;-^  +  A^if  +  A-=AV  +  T¥^=--T¥^ 

=  t\-     ^^5-  if  ^^^^  t\- 

3.  16s.  4id  =  196id     ^w5.-£(-|  of  196^)^240. 

I  mile  -=  1000  meters  ;  1  mi.  =  1000  m.  ^  |  =  1600  m.     17  mi. 
=  1600  m.  X  17  =  27200  m. ;  6  furlongs  =  200  m.  X  6  -  1200  m. ; 

82iyd.^l550j^^xi^--^75m. 
^  ^  1760         2 

27200  m.  -f  1200  m.  +  75  m.  =  28475  m.  Ans. 

27  yd.  2  ft.  9  in.  =  1005  in.  ;  17  yd.  1  ft.  11  in.  -  635  in. 
(635x1)    sq.    in.    cost    $25.40;     1    sq.    in. -=  |2540  ^  635 ; 
(1005  X  i)  sq.  in.  =  ($25.40  ^  635)  X  1005  X  i  =  $35.17f 

4.  Let  a:  =  B's  money  ;  a;+ 17.50  =  A's. 

2a:  _  a: +17.50 
5  3 

A's  rate  is  ^  of  B's;  B's  time  is  -i-J  of  A's.     To  run  the  whole 
distance,  A  needs  34  min.  ^  ij  —  36  min.     If  he  runs  2-J-  milea 


216  MANUAL    FOR  TEACHERS 

in  1G|  min.,  in  1  min.  he  runs  2^  mi.  -^-  1G|-,  and  in  36  min.  he 
runs  (2^  min.  -^  16|)  X  36  =  J  mi.  X  /^  X  \«-  =  5  mi.  Am. 

5.  x-li%    of   a;  =  96084;    ^-^=96084;   800a;-15a: 

800 

=  76867200 ;  785a;  =  76867200  ;  x  =  97920. 

6.  For  the  information  of  the  teacher,  the  following  method 
is  given  : 

V2-1_V2-1,,V2-1_2-2V2  +  1_3-2V2_3     q^/q- 
V2  +  1     V2+1     V2-1  2-1  1 

We  learn  in  algebra  that  the  sum  of  two  numbers  {x  +  y)  multi- 
plied by  their  difference  (x  —  y)  gives  the  difference  of  their 
squares  (x^  —  y').  If  the  sum  of  V2  and  1  be  multiplied  by 
their  difference  (V2  —  1),  we  obtain  the  difference  of  their 
squares  (2  —  1).  Multiplying  the  numerator  also  by  V2  —  1,  we 
retain  the  equality  and  obtain  a  divisor  that  has  no  decimals. 
See  Art.  1169,  3. 

It  is  not  expected  that  this  method  should  be  given  to  the 
pupils. 

7.  For  42  da.,  50  men  were  at  work.  To  do  the  same  work, 
30  men  would  have  required  70  da.,  or  (70  +  40)  da.  to  do  the 
whole  work.  110  da.  —  84  da.  =  number  of  days  the  contractor 
would  have  been  behindhand. 

8.  The  number  of  square  feet  in  the  wall  =  (23|  + 15|  +  23| 

-f  15|)  X  llj  =  928^  sq.  ft.     The  two  windows  contain  (19  X  5) 

sq.  ft.  =  95  sq.  ft.;  the  fireplace  contains  (4^  X  6)  sq.  ft.  =27  sq. 

ft.;  the  door  contains  (7^  X  3^)  sq.  ft.  =  26J  sq.  ft. ;  a  total  of 

95  sq.  ft.  +  27  sq.  ft.  +  26J  sq.  ft.  =  148J  sq.  ft.      There  remain 

to  be  papered  928J  sq.  ft.  -  148J  sq.  ft.  =  780  sq.  ft.  =  ^f^  sq. 

yd.     A  roll  of  paper  contains  12  X  ff  sq.  yd.;  the  number  of 

11       -11  u     4.1.      f        "780     /lov.  26\      780x1x36         ,  ., 
rolls  will  be,  therefore,  -^  -  (12  X  -j  =.  -^_^^-^;  and  ite 

$4.08X780X1X36^      Q3Q    ^,,. 
9  X  12  X  26  ^ 


NOTES   ON    CHAPTER   FOURTEEN  217 

9.    a;  +  ^=336. 

10.  The  "present  worth"  of  $365  due  in  30  da.  =  |365^ 
1.005  =  $  363.18  +  =  the  cost  of  the  horse  in  cash.  The  "  pres- 
ent worth  "  of  the  selling  price  =-  $435  ^  1.02  =  $426.47+. 
Gain=  $426.47  -  $363.18  -  $63.29,  which  is  17.43%  of  the  cost. 

If  the  seller  has  the  note  for  $435  discounted  at  a  bank,  he 
will  receive  in  cash  $435 -$8.70  -  $426.30.  If  he  uses  this 
money  to  buy  the  note  he  has  given,  he  should  pay,  at  bank  rates, 
$365-$1.82i  =  $363.17f  The  profit  would  be  $426.30- 
$363,171  =  $63,121,  which  is  17.38%  of  the  cost. 

11.  £57  Is.  Sd.  =  13700c^. ;  £2  lis.  i^d.  =  eU^d. 

13700  x^Xx  =  interest  -  ?55^  =.  616|, 

2055  a;  =  1233, 

^  =  UH ;  UU  y^-  =  7  mo.  6  da.  Ans. 

13.  A  man  that  does  only  -f  of  a  day's  work,  does  14  da.  less 
work  in  84  da.  than  the  average.  The  contractor  therefore  loses, 
in  84  da.  on  three  men,  14  da.  +  12  da.  +  9-^  da.  =  35-^  da.  He 
gains  on  two  others  10-^  da.  -f  8f  da.  =  18j^  da.  The  net  loss 
=  35-|-  da.  —  ISy^Q-  da.  =  16-g-|  da.  The  extra  17  men  have  to  do 
the  equivalent  of  16|^|-  days'  work ;  each  has,  therefore,  to  do 
16-^1-  days'  work  -v- 17  =  -||-  of  a  day's  work,  or  ^  less  than  the 
average. 

14.  Making  no  allowance  for  waste,  etc.,  two  strips,  each  260  ft. 
long,  will  be  needed  for  two  sides;  two  strips,  each  (93  —  3  —  3) 
ft.,  or  87  ft.  long,  will  be  needed  for  the  other  two,  or  520  ft.  -j-  174 
ft.  =  694  ft.  =  231^  running  yards,  1  yd.  wide,  making  231-|-  sq. 
yd.     Cost  at  90^  =  $208.20. 

The  surface  to  be  carpeted  =  (260  -  5)  ft.  by  (93  —  5)  ft.  =  85 
yd.  X  29^  yd.  Cost  =  $  2.09  x  85  X  29^  -^  f  f  =  $  4936.80. 
Total,  $4936.80  +  $  208.20  =  $  5145.  Ans. 

15.  Since  the  meeting-place  is  twice  as  far  from  A  as  from  B, 
the  first  man  goes  twice  as  fast  as  the  other ;  the  latter,  there- 


218  MANUAL   FOR   TEACHERS 

fore,  walks  2^  mi.  per  hour.     If  x  is  the  distance  between  A  and 

B,  the  first  will  require  f  hr.,  and  the  second  ^^  hr.  =  —  hr. 
5  2^  6 

— —  ^=1;  a^  =  5.     An8.  5  mi. 
5       5 

16.    Let  X  =  number  of  miles  between  A  and  C,     Then  x  —  lb 
—  distance  between  B  and  C.       i 

^i^li^  =  time   required  for    ^  15     B  x-15        ^ 

15 
first  train  to  run  from  B  to  C ;  —  =  time  required  for  second 

train  to  run  from  A  to  0.  As  the  latter  train  leaves  3  hr.  later 
and  arrives  one-half  hour  later,  the  running  time  of  the  first 
is  2^  hr.  longer. 

a:—  15  __^  I  5  . 
15     ~25'^2' 
10ar-150=:6a;-f375; 

4  a:  =  525;  a:=131i.  Am.  131 J  mi. 

1169.     2.    The  width  of  the  road  =  (60  -^- 16^)  rd. ;  its  area 
=  (104  X  60  -^  16-^)  sq.  rd.  -=  [104  X  (60  ^  16|)  -^  160]  acres. 

j^        ,      $154x104x60x2     (tjQft. 
^''  ''''  '-= 33^n60 =  * ^^^• 

n.1           ,     c        A-            $200x104      ^nf. 
The  cost  of  grading  =  - — — — =  $65. 

The  cost  of  fencing  =  $^ X  104  X  5|  -  $286. 

3.    See  Art.  1168,  6. 

V5  +  V3_.  V5  +  V3^^  V5  4- V3_5-f  2V15  +  8 
V5-V3      V5-V3      V5  +  V3  5-3 

•    ^8  +  2Vl5^^     ^ 
2 


NOTES  ON  CHAPTER  FOURTEEN  219 

V5-V3_V5- V3,,  V5  -V3_5-2Vl5  +  3 
V5  +  V3      V5  +  V3      V5-V3  5-3 

=  8^AV15  =  4-VI5. 

A  ■ 

4  +  Vl5  -  (4  -  Vl5)  =  2  Vl5  =  V60. 

4.  A  cubic  centimeter  =  .3937'  cu.  in.  =  (.3937'-^  1728)  en.  ft. 
Its  weight  =  weight  of  1  gram  =  (.3937'  X  1000  ^  1728)  oz. 
-  [(.3937'  X  1000)  -^  (1728  X  16)]  lb.  A  kilogram  -  [(1000  X 
.3937'  X  1000)  ^  (1728  X  16)]  lb.  The  weight  of  the  anchor  in 
kilograms  =  6500  lb.  -^-  the  number  of  pounds  in  a  kilogram  ;  or, 

6500  X  1728  X  16 

1000  X  .3937  X  .3937  X  .3937  X  lOOO' 

5.  :r-(a;Xyf^X^%\)  =  1500; 

a;-  — =1500;  500 a;  -  7 a;  =  750000 ; 
500 

493  a;  =  750000  ;  x  =  1521.30  -.     A71S.  $1521.30. 

The  proceeds  of  the  new  note  +  $200  must  pay  the  note  of 
$2000  ;  the  proceeds  must  therefore  be  $1800. 
^-(^XTf^xA'^)  =  1800. 

1170.   2.    (80.005  -  .013) -^  88. 

Sx 

5.  Let  x=  number  of  cents  received  by  Thomas.     Then  — 

6x  .         ^ 

=  number  received  by  Henry,  and  —  =  number  received  by 

Richard.  ^^ 

,4-^+1^4.14; 

25a;  +  15a;  +  6  a;  =103.50; 
46  a;  =103.50;   a;  =  2.25. 

6.  272  liquid  quarts  =  231  cu.  in.  X  272  -^  4.  A  dry  quart 
=  2150.4  cu.  in.  h-  32. 

231  X  272  X  32     231  x  272  x  320 


Number  of  dry  quarts 


4  X  2150.4  4  X  21504 


220  MANUAL    FOR   TEACHERS 

7.  The  tub  holds  12^  qt.  X  4|  =  b^  qt.  Both  pipes  dis- 
charge 12^  qt.  +  83  qt.  =  95^  qt.  The  time  required  to  fill  it 
=  (54J  H-  95^)  min. 

8.  The  number  of  hours  that  must  elapse  before  all  will  again 
be  together  at  the  starting  point,  is  the  least  common  multiple 
of  -^,  I",  -If.  The  least  common  multiple  of  the  numerators  is  70. 
The  smallest  fraction  that  will  contain  the  above  fractions  an  exact 
number  of  times  must  have  70  for  its  numerator,  and  for  the 
denominator  the  largest  number  that  will  divide  36,  9,  and  99 
without  a  remainder;  i.e.  the  greatest  common  divisor  of  these 
numbers.  The  G.  C.  D.  is  9,  and  the  fraction  is  ^-.  In  ^  hr., 
therefore,  A,  B,  and  0  will  be  at  the  starting  point.  A  will 
have  walked  around  the  circle  (^^s\)  56  times  ;  B,  (^  ^|)  35 
times ;  C,  (^  ^  |f )  22  times. 

Note. — The  scholars  will  readily  understand  that  the  fraction  which 
is  the  least  common  multiple  of  ■^^,  |,  and  ff,  should  have  70  for  its 
numerator.  The  following  may  make  clear  to  them  why  9  should  be  the 
denominator: 

i ,  or  -,  or  —  should  be  a  whole  number ; 

a;      36        9         99 

i.e.  —  X  — .or  -.    or  —  should  be  a  whole  number, 

a;       5         2         35 

An  examination  of  the  second  line,  in  which  the  divisors  are  inverted,  will 
show  that  70  contains  the  three  denominators,  5,  2,  and  35,  an  exact  number 
of  times ;  the  numerators,  36,  9,  and  99,  should  contain  x  an  exact  num- 
ber of  times ;  x,  therefore,  must  be  a  divisor  of  these  numbers,  etc. 

1172.  This  work  may  be  slightly  abbreviated  by  combining  the 
interest  on  the  annual  interest  into  one  item  of  6  years'  interest 
instead  of  the  three  separate  ones  of  3  years'  interest,  2  years' 
interest,  and  1  year's  interest.  In  beginning  a  new  topic,  how- 
ever, pupils  should  not  be  confused  by  short  methods. 

2.   $  1200  -f  $  300  -f  (4  -f  3  -f  2  -f  1)  years'  interest  on  $  60. 

For  partial  payments  on  notes  bearing  annual  interest,  see  Art.  1308. 
The  special  rules  for  New  Hampshire  and  Vermont  will  be  found  in  Arts. 
1309  and  1310. 


NOTES   ON    CHAPTER    FOURTEEN  221 

In  states  in  which  the  collection  of  annual  interest  is  not  allowed,  the 
teacher  should  omit  this  topic. 

1173.   In  the  older  states,  time  should  not  be  spent  upon  this 
topic. 

1176.    1.    J  of  I  a  section  =  I  of  640  A. 

2.  I"  of  ^  section  measures  80  rd,  by  160  rd. 

3.  A  line  from  the  southwest  corner  of  Sec.  1,  to  the  north- 
east corner  of  Sec.  30  (see  township  diagram  on  the  opposite  page), 
is  the  hypotenuse  of  a  right-angled  triangle  whose  perpendicular, 
the  eastern  boundaries  of  Sees.  11,  14,  and  23,  is  3  mi.  long ;  and 
whose  base,  the  southern  boundaries  of  Sees.  20,  21,  22,  and  23,  is 
4  mi.  long. 

6.  The  number  of  rods  of  fence  =  80 -f  160 -f  80  +  160 
=  480.  The  number  of  feet  =  16^  X  480  =  7920.  A  fence  4 
boards  high  requires  7920  ft.  X  4,  or  31680,  running  feet  of  boards. 
If  the  latter 
X  ^  =  15840. 


If  the  latter  are  ^  ft.  wide,  the  number  of  board  feet  ~  31680 


1184.  2.  26.50  X  .85.  4.  Multiply  135  by  69,  and  point 
off  two  places  in  the  product.  5.  Find  the  base.  6.  8.50  francs 
X  (10  X  1  X  3.25).  7.  Each  dimension  can  be  expressed  in 
decimeters,  105  X  80  X  65,  whose  product  is  the  number  of  liters  ; 
or  the  product  of  the  dimensions  in  meters — 10.5  X  8  X  6.5  — 
may  be  multiplied  by  1000.  8.  0^.75  means  .75/.,  the  denomi- 
nation in  France  being  generally  written   before  the  decimal. 

9.  1.25    marks  X  [(68  ^  10)  X  36] ;    i.e.    1^    marks  X  6.8  X  36. 

10.  The  number  of  liters  =  50  X  40  X  30  =  60000 ;  92%  of  this 
tuimber  gives  the  weight  in  kilograms  (kilos). 

1186.  These  problems  are  given  for  practice  in  obtaining  the 
approximate  values  of  the  metric  units  in  terms  of  our  weights 
and  measures.     The  use  of  39.37  in.  makes  the  work  too  tedious. 

13.    4  in.  by  4  in.  by  4  in.     A  quart  =  -2-|J-  cu.  in. 


222  MANUAL   FOR  TEACHERS 

14.  A  hectoliter  =  100  liters  =  6400  cu.  in.  -  (6400  h-  2150.4) 
bu.     6400  cu.  in.  =  (6400  ^  231)  gal. 

15.  A  liter  of  water,  64  cu.  in.,  weighs  a  kilo.  1  cu.  in.  ot 
water  =  {^^  oz. ;  64  cu.  in.  =  [(64000  -^  1728)  ^  16]  lb.  =  4000 
lb.  -^  1728. 

16.  400000000  in.  =  J  circumference.  See  Arithmetic,  Art. 
1177. 

17.  A  square  meter  =  (40  X  40)  sq.  in. 

18.  An  are  =  (400  X  400)  sq.  in.  A  hectare  =  [(400  X  400) 
X  100]  sq.  in. 

19.  Hectometer  =  100  meters  =  4000  in. 

20.  A  stere  =  (40  X  40  X  40)  cu.  in.  =  (64000  -^  1728)  cu.  ft. 

21.  1000  grams  weigh  (4000  lb.  -?- 1728) ;  1  gram  weighs  4 
lb.  ^  1728  -=  28000  grains  ^  1728. 

22.  A  kilometer  =  40000  in.;  a  mile  =  63360  in. ;  a  mile 
=  (6.336  ^  4)  Km. 

1197.  The  average  pupil  should  be  permitted  to  use  a  pencil 
for  his  first  solution  of  these  problems. 

1.  Let  X  =  the  value  of  the  second  suit.  Since  $12  and  the 
overcoat  =  2x,  the  overcoat  =  2  a;  —  12.  The  second  suit  (x)  and 
the  overcoat  (2x—  12)  =  three  times  the  first  suit  (36). 

a;-f  2a:-12  =  36;  etc. 

The  second  suit  is  worth  $  16 ;  the  overcoat,  $20.  Ans. 

2.  a; -22  +  ^^  =  ^;  etc. 

4  3 

The  arithmetical  analysis  might  assume  some  such  form  as 
this :  The  remainder  +  J  of  the  remainder,  or  -J  of  the  remain- 
der =  ^  of  original  sum.  The  remainder  =  -J-  of  original  sum  X 
|-  =  ^  of  original  sum.  The  sura  lost  is  1  —  -j^,  or  -J-J-  of  original 
sum.     As  this  is  $  22,  the  original  sum  =  $  22  X  ff  =  $  30.  Am. 


NOTES   ON   CHAPTER   FOURTEEN  223 

3.  Let   a;  =  time   past   noon;    a; +12  =  time  past  midnight. 

^^07+12.  5^^^_pi2;  4ar=12;  a;  =  3.     The  time  is  3  hr. 

5 
past  noon,  or  3  p.m.  Ans. 

4.  At  3  o'clock,  the  hour  hand  is  15  minute  spaces  in  ad- 
vance. To  be  only  5  spaces  behind,  the  minute  hand  must  gain 
10  spaces.  While  the  minute  hand  goes  1  space,  the  hour  hand 
goes  ^  space ;  so  that  each  minute,  the  minute  hand  gains  \^ 
space.  To  gain  the  10  spaces  necessary,  the  minute  hand  must 
travel  (10  ^i|-)  minutes  =  120  min.  -^  11  =  10|^  min.  The 
time  is  10^  min.  past  3. 

5.  1.  A  =  f  B;   A  =  4  B;  5  B -=  30.     B's  age  =  6  yr. ;  A's 

age  =  24  yr.  Ans. 

6.  A  takes  $  15  less  than  f  of  the  profits.  If  his  capital  is 
$30  less  than  f  of  the  whole,  the  latter  must  be  double  the  profits, 
or  $  1440.     A's  capital  =  $  525  X  2  =  $  1050  ;  B's  =  $  390.  Ans, 

Or,  A  takes  -f-|-|.  or  jf,  of  the  profits ;  he  owns,  therefore,  Jf  of 
the  capital.  If  ff  of  the  capital  +  $  30  =  f ,  or  jf ,  of  the  capital, 
^V  of  the  capital  =  $  30  ;  etc. 

80 

7 .  Let  X  =  the  number  of  sheep  ;  —  =  cost  of  each  ;  x—  5 

X 

,  .  .         2x-Vd  .  1-1     2a;-10^80 

=  number  remaining ; =  number  sold  ;  — X  —  = 

^3  3  a; 

160  a; -800  •     .       .^ 

=  sum  received  =  40. 

3a; 

160  a; -800  =  120  a;;  40  a;  =  800;  a;  =  20.  ^Im.  20  sheep. 

Or,  if  he  received  $40  for  f  of  the  remainder,  he  would  have 
received  $60  for  the  remaining  sheep.  $60  being  f  of  $80,  f  of 
the  sheep  remained,  and  \  of  them  died,  or  5  sheep.  The 
whole  number  was,  therefore,  20  sheep. 

8.  Let  X  =  A's  age ;  then  a;  +  10  =  B's  age ; 

X     a;+10      , 
-  =  — ^- —  ;  etc. 
2  3 


224  MANUAL    FOR   TEACHERS 

1198.     3.    7)  19  mi.  180  rd.  2  yd.  0  ft.  9  in. 
2  mi.  254  rd.  1  yd.  2  ft.  8|  in. 
Xl2 

33  mi.  172  rd.  0  yd.  2  ft.  1^^  in. 

4.    18  hr.  24  min.  12  sec.  -^  15  =  1  hr.  13  min.  36|  sec. 
1  hr.  45  min.  time  difference  =  1°  45'  X  15  =  26°  15'  difference 
in  longitude.     As  the  place  has  the  later  time,  it  is  more  east- 
erly. 

7.   Calling  it  1  in.  thick,  the  number  of  board  feet  =  16  X  f 
=  12.     $  40  X. 012  =  48^.     Ans. 

$  40  X  (16  X  f  X  2|)  -^  1000  =  $  1.20.  Ans. 
9.   [96  (in.)  X  90  (in.)  X  48  (in.)]  ^  2150.4. 
10.    If  the  strips  run  lengthwise,  their  number  will  be  8  yd. 
-^}  yd.  =  10J.     The  number  purchased  must  be  11,  each  9  yd 
long,  or  99  running  yards  of  carpet. 


XVIII 
NOTES  ON   CHAPTER   FIFTEEN 

While  the  work  contained  in  this  chapter  is  intended  more 
particularly  for  use  in  such  schools  as  extend  their  instruction 
beyond  the  eighth  year  of  the  elementary  course,  it  can  profitably 
replace  some  of  the  less  useful  arithmetical  topics  taught  during 
the  eighth  school  year. 

1199.  These  exercises  should  be  taken  up  without  any  prelim- 
inary explanations.  Their  previous  work  in  simple  equations 
has  so  familiarized  the  pupils  with  the  use  of  letters  to  express 
numbers,  etc.,  that  they  need  no  assistance  in  the  first  ten  exam- 
ples. The  necessary  technical  terms  should  be  employed  as  occa- 
sion requires,  and  their  meanings  should  be  made  clear ;  but 
exhaustive  treatment  of  the  diflferent  operations  should  be  left  for 
the  study  of  the  science  of  algebra  in  the  high  school. 

1200.  The  explanation  of  the  meaning  xy,  ahc,  etc.,  may  be 
deferred  until  Art.  1238.  For  the  present,  the  use  of  the  word 
coefficient  may  be  limited  to  simple  numerical  ones,  as  given  in 
the  text-book.  The  teacher  should  not  yet  explain  that  in  the 
expression  bxy,bx  may  be  considered  the  coefficient  of  y;  nor 
that  in  ^abc,  9  a  may  be  considered  the  coefficient  of  he,  and 
9  ah  the  coefficient  of  c. 

1204.  So  far,  the  pupils  have  been  required  to  add  only 
single  columns  containing  the  same  letters.  When  the  signs  are 
alike  throughout,  as  in  Art.  1199,  they  have  found  the  sum  of 
the  coefficients,  annexed  the  letter  or  letters,  and  prefixed  the 

■  225 


226  MANUAL    FOR   TEACHERS 

common  sign.  When  the  signs  are  unlike,  the  difference  between 
the  sums  of  the  coefficients  of  the  positive  and  of  the  negative 
terms  is  written,  preceded  by  the  sign  of  the  greater  sum. 

It  will  scarcely  be  necessary  to  state  to  pupils  that  algebraic  expressions 
containing  dissimilar  terms  are  added  by  placing  the  plus  sign  between 
them;  thus  the  sum  of  4a6  and  3ac,  for  instance,  is  written  4a6  +  Sac. 

1205.  The  expressions  employed  in  the  preceding  exercises 
are  called  Tnonomials,  or  algebraic  expressions  of  one  term. 
Those  of  more  than  one  term  are  called  polynomials. 

A  polynomial  of  two  terms  is  called  a  binomial;  one  of  three 
terms,  a  trinomial. 

6.  The  scholar  will  readily  see  that  in  the  addition  of  poly- 
nomials, each  column  should  contain  similar  terms;  i.e.  terms 
containing  the  same  letters.  That  the  letters  should  also  be 
afiected  by  the  same  exponents,  need  not  be  told  him  for  the 
present. 

1207.  From  some  of  the  preceding  examples,  may  be  seen  the 
use  of  the  plus  and  of  the  minus  sign  to  indicate  direction  north 
and  south,  and  east  and  west ;  past  and  future  time,  etc. 

In  2,  is  required  the  difference  between  —10°  and  -f-90°.  In 
6,  there  is  asked  the  sum  of  +40^  and  -  50^.  Calling  the  dis- 
tance north  of  the  starting  point  -\-  50  miles,  in  8,  and  the  dis- 
tance south  —  70  miles,  the  required  location  will  be  (+  60 
miles)  4-  (  —  70  miles)  ==  —  20  miles,  or  20  miles  south. 

6  and  8  are  problems  in  algebraic  addition ;  10,  like  2,  is  a 
problem  in  subtraction.  The  results  of  a  man's  transactions  dur- 
ing a  month  are  ascertained  by  subtracting  the  value  of  his  pos- 
sessions at  the  beginning  of  the  month  from  their  value  at  the 
end.  In  10,  a  man  is  worth  —$250  on  Feb.  1 ;  deducting  from 
this  +$150,  which  represents  his  condition  on  Jan.  1,  we  obtain 
-$400.  The  operation  maybe  indicated  thus:  (-$250)  — 
(-|-$150)=  —$400,  the  minus  sign  in  the  result  indicating  a 
loss. 


NOTES   ON    CHAPTER   FIFTEEN  227 

The  algebraic  analyses  of  these  problems,  if  asked  at  all,  should 
not  be  required  until  the  pupils  have  solved  them  in  their  own 
way.  The  main  object  of  teaching  subtraction  at  this  stage,  is 
to  enable  the  scholars  to  understand  the  reasons  for  the  change 
of  signs  that  accompanies  the  removal  of  a  parenthesis  preceded 
by  a  minus  sign.     See  Art.  1210. 

1210.  Considering  {a)  as  an  example  in      /  n  -ri 

fL\      OA               subtraction,  it  may  be  made      ^    '  ^. 
{o)      84                     .            '.          J                           Take  49-25 
_  .Q  ,  or      one  m  addition  (6)  by  cnang-  

ing  the  signs  of  the  subtrahend.     It  may  then  be 

written  84 -49 +  25. 

1211.  The  pupil  will  readily  ascertain  that  a  parenthesis  pre- 
ceded by  a  plus  sign  may  be  removed  without  any  alteration 
being  required  in  the  signs  of  the  quantities  enclosed  within  it. 
57  +  (33  -  16)  =  74,  may  be  written  57  +  33  -16  =  74.  In  2, 
the  signs  of  the  quantities  within  a  paren-  ^  4_  qo 
thesis  must  be  changed.  The  first  number  ^  ,  i  aq  i  op; 
within  the  parenthesis,  being  without  a  sign,  is 

QQ  positive ;  it  therefore  becomes  negative  when 

—  63  —  25  ^^^  parenthesis  is  taken  away.     The  equation 

— -— — —      then    becomes   92  —  63  -  25  =  4.     In    5,    the 

92  —  63  — ^  25 

multiplier  4  affects  only  the  quantity  within 

the  parenthesis.     The  brackets  heretofore  used,  have  been  omitted 

in  5  and  6,  to  make  these  arithmetical  equations  resemble  more 

closely  the  algebraic  equations  of  Art.  1213.     5  becomes  75  4- 

60  -  40  =  95  ;  6  becomes  75  -  60  +  40  =  55. 

1213.  It  has  not  been  considered  necessary  to  give  any  pre- 
vious practice  in  multiplying  simple  algebraic  polynomials  by  an 
ordinary  number.  The  average  pupil  will  readily  understand 
that  6  times  two  x  =  twelve  x. 

1.    12a:- 30  =  5a; +12. 

12a; -5a;  =  12 +  30;  7a;  =  42;  x  =  6. 
Proof,     6  (12  -  5)  =  30  +  12 ;  i.e.,  6  times  7  =  42. 


228  MANUAL   FOR  TEACHERS 

2.  7a;+14  =  3a;  +  50;  etc. 

3.  15  + 5a; +16  =  61;  etc. 

4.  48--3a;  =  52-4a;;  etc. 

7.  2a;-2-4a;  +  38  =  3a:-9;  etc. 

8.  12a; -30 -5a;  =  12;  etc. 

9.  5a7-12a;  +  30  =  -12;  etc. 
10.  ll-3a;+10a;=38;  etc. 

1215.     12.   3a;-3-2a;  +  4  =  12. 
13.    6a;-6-4a;  +  8-3a;  +  9  +  24  =  0. 
15.    14a; -16 -18a; -36  =  12a; +15 -6a; -12. 
Transposing,  14a;  -  18a;  -  12a;  +  6a;  =  15  -  12  +  16  +  36. 
Combining,  —  10a;  =  55. 

Changing  the  signs  of  both  members, 

10a;  =  -55. 


Or. 

X 

=  - 

-5i. 

17. 

39 
4  ' 

5a;  .a; 

4  "^2 

=¥+ 

15a; 
4    ' 

etc 

18. 

2a;: 

4 

-5- 

¥+ 

13 
6* 

19. 

3a; 

4 

f  9=2a; 

+¥- 

X 

'  i 

22. 

X  — 

20  =  ^ 

+  60. 

1216. 

16 

^  =  3. 

Multiplying  by  16 

l,3}a;- 

-7  = 

48; 

or  - 

Lla; 
3 

Clearing 

of  fractions,  11: 

i;-2] 

L  =  : 

144; 

etc. 

2. 

3a;. 

8 

=  a;-60; 

;  etc. 

7  =  4a 


NOTES   ON   CHAPTER   FIFTEEN  229 

3.    r,  +  |  +  a;  +  f  +  a;-hf  +  ^  +  f-=99. 
3  4  5  6 

6     8     3 

6.  a;  -|-  12  =  son's  present  age  ;  2  a;  +  12  =  father's  present 

age. 

rc+12  +  2a:4-12  =  138; 

3a; -138 -24  =  114; 

X  =  38,  the  son's  age  12  yr.  ago ; 

2  a;  =  76,  the  father's  age  12  yr.  ago. 

The  present  age  of  the  son  is  50  yr.  (x  + 12)  ;  the  present  age 

of  the  father  is  88  yr.  (2  a;  +  12).  Ans. 

7.  (80  +  a;)  =  2i-(60-a;);  etc. 

8.  2(ll+a;)  =  25  +  a;. 

9.  Let  X  =  the  number  of  gallons  originally  in    the  cask. 

X  3  a;.  3  a; 

-  =  amount  drawn  off,  leaving  —  in  the  cask.    60 —  =  num- 

4  ^4  4 

ber  of  gallons  required  to  fill  the  cask. 

24  =  60  -  5;^  ; 
4 

96  =  240 -3a;;  etc. 


10. 

1-40  = 

104. 

11. 

a;4-430_ 

X 

=  4  + 

76 

X 

Clearing  of  fractions,  x  +  430  =  4  a;  -{-  76. 
Transposing,  a;  —  4  a;  =  76  —  430. 

Combining,  —  3  a;  =  —  354. 

Changing  the  signs  of  both  members, 

3  a;  =  354. 
Or,  x=  118,  the  smaller  number; 

X  +  430  =  548,  the  larger  number. 


230  MANUAL   FOR  TEACHERS 

12.  Let  X  =  the  number  of  $2  bnls. 
Then  29  -  x  =  the  number  of  $  5  bills. 

2a; +  5 (29 -a;)  =  103;  etc. 

13.  4(a;4-4)  =  a:  +  34. 

14.  (a;  +  3)X  — =  225. 
Multiplying,      '  M£±^  =  225. 

X 

Clearing  of  fractions,    180a;  +  540  =  225  x ;  etc. 

15.  33(a;+l)  =  40a;+12. 

16.  The  numbers  are  x  and  x  -\- 17. 
Then  a;  +  a;+ 17  =47;  etc. 

17.  Let  X  =  number  of  years. 

The  mother's  age  will  then  be  41  +  x,  and  the  son's  5  -f  a;, 

5  +  a;=i(41+^)  =  ^^; 
15  +  3a;  =  41  +  a;;  etc. 

1218.  1.    Substituting  the  value  of  8  a;,  the  equation  becomes 

16  +  7y  =  44;  etc. 
2.    9  +  52  =  34;  5z  =  25;  z  =  5.  Ans. 

1219.  11.  3a;+14y=78 

2a;+14y  =  66 

Subtracting,         x  =12 

Substituting  this  value  of  x  in  the  first  equation, 

36  +  14  2/  =  78 ;  etc. 

14.  Multiplying  the  first  equation  by  2,  we  have  2x-\-2y 
=  30.  Subtract  the  new  equation  from  the  second  one,  2  a;  -f  3  y 
=  38,  thus  finding  the  value  of  y.  Substitute  this  value  in 
either  of  the  original  equations. 


NOTES  ON  CHAPTEH  FIFTEEN  231 

17.  Multiplying  the  first  equation  by  3,  and  the  second  by  2, 
we  have  6  a;  +  9  y  =  120  and  6  ^  +  4  y  =  70. 

18.  Multiply  the  first  by  2,  and  the  second  by  7. 

19.  Multiply  the  first  by  9,  and  the  second  by  5. 

20.  Multiply  the  first  by  8,  and  the  second  by  3. 

1221.  21.  If  we  add,  we  have  2.t  =  22.  Or,  subtracting, 
2?/^=  14.  By  adding,  y  is  eliminated;  by  subtracting,  x  is 
eliminated.  From  this  example,  pupils  should  see  that  either 
addition  or  subtraction  may  be  employed  to  eliminate  one  of  the 
unknown  quantities ;  and  that  either  of  the  two  may  be  elimi- 
nated, as  may  be  found  convenient. 

27.    hx-^y^      5,  28.    3:r4-5y-=-8, 

3a;-5?/--4.  2a;-    ?/ =    12. 

29.  -  10  re +  2/  =  -  1,     (1)         Subtract  (1)  from  (2). 
-    5a:  +  2/=      9.     (2) 

30.  3a;  +  8y-204,  31.    3:r  +  2?/-252, 
10a:  +  5y-160.  7a:  +  5y---609. 

34.  3.r  +  7-15y-20;     3:r  -  15?y  =-  -  27.     (1) 
7a;- 6 --10?/+    G;     7a;-10y-      12.     (2) 

Multiply  (1)  by  7  and  (2)  by  3,  to  eliminate  x\  or  (1)  by  2 
and  (2)  by  3,  to  eliminate  y. 

35.  51  a; +  44  2/ -804,    (1)  Multiply  (1)  by  8,  and  (2) 

45a;-32y=    72.    (2)  by  11.     Add. 

37.      2:c  -  22  -  2  ?/  +  18  =  6,  The  pupils  should  be  taught 

15  re -I- 135  =  32  •?/ —  96      ^*^  indicate  the  common  de- 
nominator of  2/  —  3  and  15, 
as  15  {y  —  3),  which  contains  2/  —  3,  15  times  ;  and  15,  (y  —  3) 
times. 

39.   2re  +  52/  +  3  =  18rc-242/-12;  -  16a;+ 29y  =  -  15. 
4a;-73/  +  5  =  5a;-102/  +  10;  -x-^Zy  =  h. 


232  MANUAL   FOR  TEACHERS 

1222.  1.      a:  +  2/  =  37  ;         This  problem  can  also  be  solved 

2a;-f-3y  =  96.      ^7  tbe  use  of  one  unknown  quan- 
tity, by  calling  the  numbers  x  and 
37  —  X.     The  equation  becomes  2a;  +  3  (37  —  a;)  =  96  ;  or,  2a:  + 
111 -3a;  =  96. 

2.  Using  one  unknown  quantity,  the  numbers    are  x  and 
a;  +  28.     5  a;  -  2  (a;  +  28)  =  197. 

3.  5a;  +  3y  =  37;  6a;-5y=10. 

4.  a;-fy  =  65;  a;  — 2/  =  19. 

By  one  unknown  quantity,  a;  +  (a;  -f  19)  =  65. 

5.  a;  -f  2/  =  32 ;  2a;  +  Sy  =  103.     See  Art.  1216,  12. 

6.  a;  +  y  =  25;  7a;  +  5y=145. 

7.  10a;-f  42/  =  38;  6a;  +  7y  =  32. 

8.  5a; +  3y  =  375  (cents);  8a; 4-y  =  505  (cents). 

9.  125  (a;)  +  45  (4a;)  +  10  (8a;)  +  5  (i  a;)  =  1550. 

11.  a;  +  y=19;  y+10a;-(a;  +  10y)  =  45. 

12.  13a;  =  5y;  a;  +  y=126. 

13.  15a;  =  83/;  a;  — y  =  — 147.  -^  being  2i,  proper  fraction, 
any  equivalent  fraction  must  have  a  denominator  exceeding  the 
numerator. 

1223.  4.  Eliminate  2  by  comparing  the  first  and  the  sec- 
ond equation,  multiplying  the  latter  by  5.  Multiply  the  second 
by  3  and  compare  with  the  third  equation,  eliminating  z. 

5.  First  eliminate  y. 

6.  a;  +  a;  +  y=42;  3a;  + 3y  -  a;  +  y  =  96. 

7.  15a;-25y  +  30  =  4a;  +  2y;  11a;  — 27y  =  -  30. 
96-3a;  +  6y  =  6a;-f  4y;  -9a;  +  2y  =  — 96. 

9.    15a;~9-9a;  +  57  =  24-6y  +  2a;, 

16a;  +  8y  -  18a;  -f  14  =  12y  +  36  -  4a;  -  5y. 
Transposing  and  combining,  4a;-{-6y  =  —  24;  2a;-j-y  =  22. 
Eliminate  x  by  dividing  the  first  equation  by  2 ;  etc. 


NOTES   ON   CHAPTER   FIFTEEN  233 


2.  Let  X  =  number  of  B's  chestnuts ;  a:  +  18  =  number  of  As 
chestnuts. 

rr+18  +  4  =  4(a;  +  4). 

3.  rt'  +  y  =  8;  23a;  +  ITy  =  166. 

4.  a;  +  y  =  55;  a;  +  z  =  62  ;  3/  +  z  =  83. 

Comparing  the  first  two,  we  get  y  —  z  =  —  l  \  adding  this  to 
the  third  eliminates  z ;  etc. 

«•   ^  =  1+^  +  1+24- 

7.   .-1  =  510.      8.   .-^=147. 

10.  4^.=  f+16. 

7        5 

11.  Let  a:  =  value  of  the  clothes,     a:  +  280  =  yearly  wages. 
-J  {x  +  280)  =  wages  for  6  months  =  x -{-  130. 

Clearing  of  fractions,  x  +  280  =  2x-\-  260  ;  etc. 

1228.     14.  3  a: +  6 

2a;  — 3 


2a;(3a;  +  6)     6a;^4-12a; 
-3(3a;+6)  -9a;-18 

6a;"' +  3a; -18  Ans. 

1236.     5.    5a;'^  +  85-3a;^  +  63  =  198;  etc. 

7.  a;2  +  2a;+l-a;'^-49;  etc. 

8.  43/^  +  20  — 63/^+54  =  24;  etc. 

9.  Clearing  of  fractions.  Art,  1221,  37,  we  have  (z  -f  7)  (z  —  9) 
:(z-5)(z-3). 

Performing  the  multiplication  indicated, 

z^-2z-63  =  z'-8z+15; 
6  z  =  78  ;  z  =  13.  Ans. 


234  MANUAL   FOE  TEACHERS 

10.     20x(x-{-l)  =  S0x(x-  1)  ;  etc.    Divide  by  x. 
13.    (a;  +  4)(a;  +  4)  =  8a;  +  80;  etc. 

15.  6A-^  +  36  =  5a;^+72. 

16.  x"^—  Qx-}-9  —  (x^—lOx-l-  25)  =  12 ;  removing  the  paren- 
thesis. x'-6x-i-d-x'-j-l0x-2b=12;  etc. 

18.  The  common  denominator  is  36  a:.     Clearing  of  fractions, 

9a;' +144  =  4a;' +  324;  etc. 

19.  (a:+7)(.'r  -  9)  =  (a;-3)(a;  — 5). 

20.  (2/-9)(2/  +  7)  =  (2/-3)(y-6). 

1237.     1.    Let  X  --=  the  breadth  ;  2x  =  the  length.     The  area 
=  a;x2a;-2a;'  =  1800;  etc. 

2.  Let  X  =  the  length  of  one  edge.     The  area  of  one  face 
=  x^ ;  that  of  six  faces  is  6a;',  and  is  equal  to  96  sq.  in.  \ 

6a;' =  96;  etc. 

3.  Let  X  =^  one  number  ;  —  =  the  other. 

5 

xx^=^  =  their  product  =  80 ;  etc. 
5         5 

100      100       25 

6.   40%    of  a   number  (a;)  =  ^ ;    30%  of  ^  =  :i  of  ^' 
Q       Q  5  5        10         5 

.|.    1  =  300;  etc.  , 

Let  X  =  the  length  of  the  perpendicular ;  -^  =  the  length 


of  the  base.     The  area  =  -I x  X  —-]—  — --• 

2V       4y     8 

§^=96;  3ar'  =  768;  a:"  =  256;  a:=zbl6. 
8 

Neglecting   the  negative  result,  the   perpendicular   measures 


NOTES   ON   CHAPTER   FIFTEEN  235 

16  rd.,  and  the  base  12  rd.     The  hypotenuse  =  Vl6"'  +  12'  rd. 
-  20  rd. 

8.  x'  +  (— )'=  15' ;  x'  +  ^=  225 ;  16a;'  +  9^;'--=  3600 ;  etc. 

9.  (.7; +  9)'  =  a;' +  15';  a;' +  18a;  + 81  =  a;' +  225  ;  etc. 
10.    (a;+l)'-a;'  =  49;  etc. 

1238.  The  pupils  should  be  informed  that  the  product  of  the 
numbers  represented  by  two  letters  is  represented  by  writing 
the  letters  together  ;  thus  a  times  b  is  written  ah,  m  times  n  is 
written  mn,  just  as  3  times  x  is  written  3^7. 

1242.     1.    a;' +  60; +  9.  6.    x^-\-2x-\-l. 

2.  a;'-12a;  +  36.  7.    :r'-4a;  +  4. 

3.  ^'-8^  +  16.  8.    a;' -10 a; +  25. 

1244.     1.    a:'  +  6:t'+9  =  49;  ^  +  3  =  ±7.  Ans. 
2.    a;'  -  12:r  +  36  =-  64  ;  a;  -  6  =  ±  8.  Ans. 

5.  a;^  +  18a;  +  81  =  19  +  81  =  100;  .t  +  9  =  ±10.  Ans. 

6.  a;'  +  2a-+l  =  24  +  l  =  25;  a:  +  I  =  ±  5.  Ans. 

7.  a;'- 14a; +  49  =  15  + 49;  a;— 7  =  ±8.  Ans. 

1246.  1.    a;'-6a;  +  9  =  7  +  9;a;-3  =  ±4;  etc. 

2.  a;' -12a; +  36  =  108 +  36;  etc. 

3.  a;'+2a;  +  l  =48  +  1;  etc. 

1247.  The  first  member  is  made  a  complete  square  by  adding 
the  square  of  -J-  of  the  coefficient  of  x. 

1248.  1.    a;'  +  .r  +  i  =  12  +  i  =  4^. 

a;  +  -i-  =  ±|-;  a;  =  f,  or— f  =  3or—  4.  Ans. 

2.  a;^-3a;  +  f  =  10  +  1  =  ^; 

.T-f  =  ±1-;  etc. 

3.  ^.  +  5^_|_(5)2__4_^(^5y.  etc. 


236  MANUAL   FOR  TEACHERS 

1249.  1.   x'-x  =  6',  x^~x  +  i  =  Q-\-i  =  ^;  etc. 

1250.  1.    12a; -a;' -32. 
Changing  the  signs,  and  rearranging, 

a:' -12a;  =  -32. 
Completing  the  square,  a;'  —  12  a;  -f-  36  =  -  32  +  36  =  4. 
Extracting  the  square  root,  a;  —  6  =  rb  2. 

Transposing,     a;  =  -|-2  +  6  =  8;  ora;  =  —  24-6==4. 
12 -a;  =  12-8  =  4;  or  12  -  4  =  8. 

8  and  4,  or  4  and  8.  Ans, 

2.  a;^  + 50a; +  625  =  2400 +  625  =  3025. 

a;  +  25  =  =h55; 

a;  =  30  or  -  80. 
Neglecting  the  negative  result,  the  altitude  is  30  ft.  Ans. 

3.  a;2  +  225  +  30a;  +  a;'  =  5625. 
Transposing,        2a;'  +  30a;  =  5625  -  225  =  5400. 
Dividing  both  members  by  2,       x'  +  15  a;  =  2700. 
Completing  the  square,    a;'  +  15  a;  +  (^y  =  2700  +  (-»/)' ;  etc. 

4.  Perpendicular  =  V^a;'  — a;'  =  V^^  =  f  a;. 

Area  4(.xL^)  =  3^  =  150. 

Clearing  of  fractions,  3  a;'  =  1200, 

a^=   400, 
a;  =     20. 

Base  =  20  yd. ;  hypotenuse  =  20  yd.  x  1 J  =  25  yd.  Avs. 

5.  The  convex  surface  =  6  (a; +  a;  +  a;  + a;)  =  24  a;;  the  surfnce 
of  the  two  bases  =  2  a;' ;  the  entire  surface  =  2a;'  +  24a;  =  170 ;    j 
5;"  + 12a;  =  85;  etc.  I 

6.  The  area  of  the  walk  =  area  of  outside  rectangle  —  area 
of  inner  rectangle. 


NOTES   ON    CHAPTER    FIFTEEN  237 

(40  +  2:c)(50+  2a;)  -  40  X  50  -  784  ; 
2000  -{-180x  +  4:x'~  2000  =  784 ; 
4a;^+ 180  a;  =  784; 
ar'^- 45a;  =196;  etc. 

7.  12  acres  =  1920  sq.  rd. 

i^'  =  1920  ;  15a;'  -  15360  ;  x'  =  1024  ;  a;  =  ±  32. 
8 

Base  =  32  rd. ;  perpendicular  =  32  rd.  X  If  =  60  rd.  ;  hypote- 
nuse =  V32H^~60'  rd.  =  68  rd.     Diagonal  =  68  rd.  Ans. 

8.  a;' +  30' =(50 -a;)'. 

x"  +  900  =  2500  -  100  a;  +  a;* ; 
100  a;  =1600;  a;  =  16. 
AC=  16  ft.,  CB,  the  part  broken  off  =  50  ft.  -  16  ft.  =  34  ft.  Ans. 
Or,  making  BC=  x,  AC=  50  —  x. 
Then,  (50 -a;)' +  30'  =  a;'. 

2500  -  100  a;  +  a;'  +  900  =  a? 
-  100a;  =  -  3400  ; 

X  =  34,  the  length  in  feet  of  the  part  broken  off. 

9.  60' +  (58  -  a;)' =  56' +  a;' ; 

3600  +  3364  -  116  a;  +  a;'  =  3136  +  a;' ; 
a;'  -  a;'  -  116  a;  =  3136  -  3600  -  3364 

-  116  a;  =  -  3828,  or  116  a;  =  3828  ; 
a;  =  33  =  AK 

The  length  of  the  ladder  in  feet  =  V56'  +  33'  =  V3136  +  1089 
=  V4225.     65  ft.  Ans. 

10.    From  ABB,  the  square  of  BB  =  13'  -  (15  -  a:)'.     From 
BOB,  the  square  of  BB  =  A'-  x\     Therefore 

13' -(15 -a;)' =  4' -a;'; 
or,  169  -  (225  -  30  a;  +  a;')  =  16  -  x\ 


238  MANUAL   FOR  TEACHERS 

Removing  the  parenthesis,  169  —  225  -{-S0x  —  3^=  16  —  a:*. 
Transposing  and  combining,  30  a;  =  16  -  169  +  225  =  72 ; 

^  =  ^. 

^i)  =  V^C-(7Z)-V4'-2f  =  Vl6-J^  =  J^  =  3f 

Altitude  =  3J  ft.  Ans. 

11.    AF=-  \^AB'  -  BF"  =  V1156  -  256  =  V9'00  =  30  ; 
FC=  -y/BC  -  BF'  =  V400  -  256  =  Vl44  -  12 ; 
AC=  AF+  F0=  30  +  12  =  42.     42  ft.  Am. 
Let  AE=x]  FC=4:2-x; 

FD'  =  AD'  -  AF'  =  26'  -  x'  =  676  -  x' ; 
FD'  =  DC  -  FC^  =  40'  -  (42  -  xf  =  1600 
-(1764-8457  + a;'). 
Therefore      676  -  a;'  =  1600  -  1 764  +  84  a:  -  a;* 

-  84  a;  =  1600  -  1764  -  676  =  -  840; 

X  =  10. 
ED  ==  V26'  -  10'  -  V576  -  24  ; 
or.  =  V40'-32'  =  V576  =  24.     24  ft.  Ans, 


XIX 

NOTES  ON   CHAPTER  SIXTEEN 

The  geometry  work  contained  in  this  chapter  should  be  com- 
menced not  later  than  the  seventh  year  of  school,  and  should  be 
continued  throughout  the  remainder  of  the  grammar-school 
course. 

1251.  No  formal  definitions  of  lines,  angles,  etc.,  should  be 
given  at  the  beginning.  After  drawing  angles  of  various  sizes 
and  with  lines  of  different  lengths,  the  pupils  will  be  able  to 
understand  that  "  an  angle  is  the  difference  in  direction  of  two 
straight  lines  that  meet  in  one  point,  or  that  would  meet  if 
produced." 

1255.  The  semi-circular  protractor  is  better  than  the  com- 
mon rectangular  one  for  beginners,  as  they  see  more  clearly  by 
using  the  former  that  an  angle  is  measured  by  the  arc  of  a  circle. 
Two  protractors  are  printed  on  a  fly-leaf  in  the  back  of  the  text- 
book, for  the  use  of  such  pupils  as  cannot  procure  others.  Pro- 
tractors made  of  stout  manilla  paper  can  be  obtained  from  the 
Milton  Bradley  Co.,  New  York,  at  one  cent  each  in  quantities. 
A  large  protractor  is  needed  for  blackboard  use.  This  can  be 
made  of  pasteboard ;  or  wooden  ones  can  be  bought  of  the 
Keuffel  &  Esser  Co.,  New  York. 

Many  scholars  that  are  able  to  measure  an  angle  one  of  who.se 
sides  is  horizontal.  Fig.  1,  find  it  difficult  at  first  to  ascertain  the 
number  of  degrees  in  an  angle  formed  by  two  oblique  lines, 
Figs.  2  and  3.    They  should  be  permitted  to  discover  the  method 

239 


240  MANUAL    FOR   TEACHERS 

for  themselves.  All  that  is  necessary,  is  to  place  the  center  (A) 
of  the  base  of  the  protractor  on  the  vertex  of  the  angle,  Figs.  1-3, 
and  the  edge  of  the  protractor  on  one  of  the  sides,  the  other  side 
cutting  the  circumference.  In  Figs.  2  and  3,  the  number  of 
degrees  in  the  angle  JCAZ  is  determined  by  the  number  of  degrees 


in  the  arc  BM,  and  the  upper  row  of  figures  is  used,  having  thv* 
zero  mark  at  B.  In  Fig.  1,  the  number  of  degrees  in  the  angle 
is  measured  by  the  arc  CM,  which  requires  the  use  of  the  lower 
row  of  figures. 

1256.  The  average  class  will  find  the  100  exercises  to 
Art.  1269,  inclusive,  sufficient  for  the  first  year's  work.  This 
will  give  three  per  week,  and  leave  some  time  for  review.  Pupils 
should  work  the  exercises  at  home  without  any  preliminary 
discussion  in  class.  After  the  exercises  are  brought  in,  they 
should  be  done  on  the  blackboard,  at  which  time  the  mistakes 
made  can  be  pointed  out.  While  first-class  drawing  cannot  be 
expected  from  the  instruments  used  by  school-children,  the 
teacher  should  exact  the  best  work  possible  under  the  circum- 
stances. A  hard  pencil,  kept  sharp,  is  necessary  to  secure  the 
requisite  fineness  of  line. 

1.    In  drawing  an  angle,  commence  at  the  vertex. 

This  exercise  is  given  to  remove  the  impression  sometimes 
formed,  that  the  size  of  an  angle  depends  upon  the  length  of  the 
lines,  instead  of  their  greater  or  less  difference  in  direction. 

In  this  and  all  other  exercises,  the  pupils  should  be  encouraged 


NOTES   ON   CHAPTER    SIXTEEN  241 

to  commence  occasionally  with  an  oblique  line.  No  two  results 
should  be  exactly  alike.  If  two  pupils  compare  notes,  it  should 
be  for  the  purpose  of  producing  a  different  drawing.  One 
pupil's  angle  may  have  its  vertex  at  the  right,  another  at  the 
left ;  one  vertex  may  be  above,  another  below  ;  etc.  The  better 
the  teaching,  the  greater  will  be  the  variety  of  results  in  exer- 
cises that  permit  of  variety. 

3.  It  is  expected  that  the  pupils  will  see  for  themselves  that 
each  arc  will  contain  ^  of  360°. 

4.  Using  the  ruler,  draw  the  first  line  of  any  convenient 
length  and  in  any  direction.  Placing  A  of  the  protractor  at 
either  end,  mark  off  45°,  being  careful  to  use  the  proper  row 
of  figures.  Remove  the  protractor ;  place  the  ruler  so  that  its 
edge  just  touches  the  end  of  the  line  and  the  45°  point,  and  draw 
the  second  line.  This  latter  should  not  be  of  the  same  length  as 
the  first,  unless  for  some  good  reason ;  so  that  pupils  will  not 
consider  that  the  lines  forming  an  angle  should  be  equally  long. 

Write  the  number  of  degrees  in  each  angle. 

5.  The  teacher  should  not  inform  the  pupils  in  advance  how 
many  degrees  they  will  find  in  the  second  angle.  They  should 
measure  it  for  themselves,  using  the  protractor. 

In  drawing  these  angles,  the  figure  in  the  book  should  not  be 
followed.  The  second  line  should  be  drawn  to  the  left  in  some 
cases  ;  the  lower  angle  may  be  made  60°  ;  etc. 

When  two  lines  meet  to  form  two  angles,  it  is  not  at  all 
necessary  that  the  point  of  meeting  should  be  at  the  center  of  one 
line. 

1257.  Pupils  should  be  taught  that  horizontal  lines  are  lines 
parallel  to  the  surface  of  still  water.  Floating  straws  are  hori- 
zontal, and  may  point  in  any  direction.  A  spirit  level  is  used 
by  the  carpenter  to  determine  whether  or  not  a  beam,  for 
instance,  is  horizontal.  A  vertical  line  is  one  that  has  the  direc- 
tion of  a  plumb  line,  which  is  used  by  a  mason  to  ascertain  if  a 
wall  is  perpendicular. 


242  MANUAL   FOR   TEACHERS 

In  drawings,  however,  lines  that  will  be  horizontal  when  the 
paper  is  placed  upon  the  wall,  are  called  horizontal  lines  ;  and 
lines  that  will  be  vertical  when  the  paper  is  placed  upon  the 
wall,  are  called  vertical  lines. 

6.  The  perpendicular  need  not  be  drawn  to  the  center  of 
either  of  the  others,  nor  need  it  always  be  drawn  above.  The 
teacher  should  encourage  variety. 

8.  The  pupils  should  draw  these  lines,  and  mark  in  each 
angle  the  number  of  degrees  it  contains.  Encourage  the  greatest 
possible  variety  in  the  size  of  the  angles  and  the  direction  of  the 
lines. 

9.  While  pupils  may  be  able  by  this  time  to  give  the  result 
without  drawing  the  angles  and  measuring  the  second  one,  the 
teacher  should  not  fail  to  give  them  the  necessary  practice  in 
constructing  angles  of  a  given  number  of  degrees,  and  in 
measuring  the  contents  of  others. 

Many  scholars  make  as  ridiculous  mistakes  in  the  measure- 
ment of  angles  as  they  do  in  their  work  in  numbers,  frequently 
reading  the  wrong  figures,  and  figure?  from  the  wrong  row  — 
marking  an  angle  of  45°,  for  instance,  135° ;  etc.  They  should 
learn  to  **  approximate"  the  size  of  an  angle,  as  well  as  to  "  esti- 
mate "  the  probable  answer  to  an  arithmetical  problem.  An 
acute  angle  should  not  be  marked  as  containing  over  90° ;  etc. 

10.  Having  learned  by  observation  that  the  sum  of  two  adja- 
cent angles  is  180°,  the  pupils  should  now  discover  that  the  sum 
of  any  number  of  angles  formed  on  one  side  of  a  straight  line  is 
180°.  When  they  have  learned  this  from  drawing  the  first 
exercise,  they  may  be  permitted  to  calculate  the  result  in  the 
other  two,  especially  as  the  protractors  are  not  marked  for 
fractions  of  a  degree. 

The  first  exercise  should  show  the  same  variety  in  the  work  of 
the  diflferent  pupils  as  has  been  recommended  for  previous  work. 

12.  In  constructing  a  square,  the  protractor  is  used  to  erect  a 
perpendicular  at  each  corner.     These  perpendiculars  are  made 


NOTES   ON   CHAPTER   SIXTE*«bi^S.-*-**^      243 

equal  to  each  other  and  to  the  original  line.  A  fourth  line  is 
drawn.  The  accuracy  of  the  work  may  be  tested  by  measuring, 
with  the  protractor,  the  two  upper  angles. 

The  base  lines  used  by  different  pupils  should  be  of  different 
lengths.  The  pupils  should  be  permitted,  also,  to  construct  the 
square  in  their  own  way.  Some  may  erect  a  perpendicular  at 
one  end  of  the  given  line,  and  at  the  extremity  of  the  second 
line  erect  another  perpendicular.  Some  pupils  may  not  measure 
the  first  line,  drawing  the  second  and  third  lines  lightly  of  indefi- 
nite length,  and  using  compasses  to  make  them  equal  to  the  first. 
In  this  case,  the  light  lines  should  not  be  erased  ;  but  the  square 
should  be  marked  off  by  heavier  lines. 

It  is  a  good  practice  to  have  the  pupils  give  a  written  descrip- 
tion of  their  method  of  working  one  of  these  exercises,  which 
should  be  accepted  as  a  regular  composition.  The  language 
should  be  correct;  the  proper  technical  terms  should  be  em- 
ployed ;  and  there  should  be  sufficient  detail  to  enable  any  one  not 
familiar  with  the  work  to  understand  just  how  it  was  done. 

13.  Pupils  should  be  permitted  to  learn  for  themselves  from 
this  exercise  and  from  14,  that  vertical,  or  opposite,  angles  are 
equal. 

17.  After  drawing  the  required  lines,  the  scholar  should  mark 
in  each  angle  its  contents  in  degrees. 

19.  This  exercise  should  enable  the  pupil  to  see  that  the  sum 
of  all  of  the  angles  formed  about  a  point  will  be  360°. 

20.  The  teacher  should  not  give  unnecessary  assistance.  If  the 
scholars  have  a  few  days  in  which  to  work  out  an  exercise,  they 
should  find  no  difficulty  in  managing  this. 

The  word  "  adjacent "  in  geometry  is  applied  to  each  of  the 
two  angles  formed  by  one  straight  line  meeting  another.  In  19, 
the  two  lower  angles  are  adjacent ;  but  none  of  the  upper  three 
angles  is  adjacent  to  any  other,  because  three  straight  lines  are 
used  to  construct  two  angles  in  each  case.  No  two  angles  in  20 
are  adjacent,  and  no  two  are  vertical. 


244  MANUAL   FOR  TEACHERS 

21.  There  are  no  adjacent  angles.  They  are  all  vertical, 
because  the  lines  forming  each  are  produced  to  form  an  opposite 
angle. 

22.  The  pupil  is  not  yet  ready  to  do  this  in  the  geometrical 
way.  The  teacher  should  be  satisfied  if  he  adds  65°  and  25°, 
and  uses  the  protractor  to  make  an  angle  of  90°  ;  etc. 

24.  If  the  pupil  examines  a  clock,  he  will  see  that  the  num- 
ber of  degrees  between  12  and  1  is  ^^  of  360°,  or  30°.  He  has 
learned  already  that  the  length  of  the  sides  has  nothing  to  do 
with  the  magnitude  of  the  angle. 

25.  The  minute  hand  goes  90°  in  a  quarter  of  an  hour.  The 
hour  hand  goes  30°  in  an  hour ;  15°  in  \  hr. ;  7-J-°  in  \  hr. 

To  ascertain  the  angle  at  12 :  15,  the  pupils  should  draw  a  clock 
face,  locating  the  hands  properly.  Some  will  place  the  hour 
hand  at  12,  forgetting  that  it  has  gone  7^°  in  \  hour;  and 
will  give  the  answer  as  90°  instead  of  the  correct  one  of 
90°  —  7^°,  or  82°  30'.  At  6  :  30,  the  minute  hand  is  at  6,  and 
the  hour  hand  is  half  way  between  6  and  7,  or  -J-  of  30°  =  15°. 
At  8 :  20,  the  minute  hand  is  at  4,  and  the  hour  hand  J  of  the 
way  between  8  and  9  —  the  number  of  hour  spaces  being  4-J-, 
corresponding  in  degrees  to  30°  X  4 J  =  130°. 

Pupils  should  understand  that  while  the  angle  at  4  o'clock  is 
30°  X  4,  or  120°,  and  while  the  angle  at  5  o'clock  is  30°  X  5,  or 
150°,  the  angle  at  7  o'clock  is  not  30°  X  7,  or  210°.  By  making 
a  drawing,  they  will  see  that  in  the  last  case  the  angle  should  be 
measured  on  the  left,  which  will  make  it  30°  X  5,  or  150°. 

Note.  —  Angles  of  180°  210°,  etc.,  may  be  left  for  more  advanced  work. 

1260.  From  26  should  be  learned  that  two  lines  perpendicular 
to  a  third  line  are  parallel  to  each  other ;  and  from  27,  that  two 
lines  running  in  the  same  direction  and  making  the  same  angle 
with  a  third  line,  are  parallel  to  each  other. 

29.  DE  will  be  drawn  parallel  to  BC  by  means  of  the  pro- 
tractor, an  angle  of  58°  being  made  at  the  intersection  of  A£  and 


NOTES   ON   CHAPTER   SIXTEEN 


245 


Fig.  4. 


DE.  Six  of  the  twelve  angles  will  contain  58°  each ;  the 
remaining  six  will  each  measure  180°  —  58°  —  122°.  The  pupils 
should  be  permitted  to  ascertain  this  for  themselves. 

The  card  suggested  in  the  note  is  to  be  used  in  schools  in  which 
small  wooden  triangles  are  not  obtainable. 

32.  To  draw  from  P,  a  line  parallel  to  the  oblique  line  AB, 
place  the  perpendicular  of  the 
triangle  on  the  line  AB,  Fig.  4, 
and  place  the  ruler  XY  against 
the  base  of  the  triangle.  Hold- 
ing the  ruler  in  position,  slide 
the  triangle  along  it  until  the 
perpendicular  passes  through  the 
point  P.  A  line  BE  drawn 
along  this  side  of  the  triangle 
will  be  parallel  to  AB. 

Time  should   be  given   the  pupils 
to  discover  this  or  a  similar  method  of  drawing  by  means  of  a  ruler  and  a 
triangle  a  line  parallel  to  another  line.     The  method  may  be  made  the  sub- 
ject of  a  composition. 

33.  QR  and  C/Fare  drawn  parallel  by  means  of  the  ruler  and 
the  triangle.  They  may  lie  in  any  position,  care  only  being  taken 
to  cut  them  by  a  line  making  angles  of  50°  and  130°  with  one  of 
them.  Three  of  the  remaining  six  angles  will  measure  50°  each ; 
and  the  others,  130°  each. 

1261.  35.  If  the  work  is  done  as  it  should  be,  the  angle  at 
C  will  measure  80°. 

The  line  AB  does  not  need  to  be  horizontal ;  nor  should  all  the 
pupils  draw  AB  of  the  same  length. 

36.  The  third  angle  will  measure  60°. 

37.  There  are  68°  in  a,  and  57°  in  c.  In  5,  there  are  180° 
—  (68°  +  57°),  or  55°.    In  d,  which  is  vertical  to  h,  there  are  55°. 

38.  The  angle  e  should  measure  28°,  and/ 120°.  There  are 
180°  in  e  (28°)  +  g  +/(120°). 


246  MANUAL   FOR  TEACHERS 


39.  PRQ  =  180°  -  (70°  -f  60°)  -  50°.  PES  =  180°  -  50° 
=  130°.  PR8  is  therefore  equal  to  the  sum  of  the  angles 
P  and  Q. 

40.  180°.  Am. 

41.  180° -(36° +  65°). 

45.  In  measuring  the  side  of  a  triangle,  use  the  smallest  frac- 
tion marked  on  the  ruler.  When  the  ruler  in  use  has  the 
denominations  of  the  metric  system  on  one  face,  that  face  should 
be  used,  and  the  length  of  the  line  given  in  millimeters.  This 
will  not  require  any  teaching  of  the  metric  system  beyond  show- 
ing pupils  how  to  read  their  rulers,  and  it  will  do  away  with  the 
need  of  using  fractions. 

46.  The  length  of  each  side  should  be  marked.  If  two  of 
them  are  not  found  exactly  equal,  the  construction  is  faulty. 

47.  The  three  sides  should  be  equal. 

50.    Each  of  the  oblique  angles  will  contain  45°. 

52.  Angles  2  and  3,  90°  each  ;  4,  50°  ;  5  and  6,  40°  each. 

53.  Angles  1  and  4,  67^°  each  ;  2  and  3,  90°  each ;  5  and  6, 
22^°  each. 

54.  The  angle  ;?  contains  120°;  m,  30°;  n,  30°. 

1266.  56.  Construct  this  parallelogram  by  drawing  two 
lines  of  the  given  lengths,  meeting  at  an  angle  of  60°.  By 
means  of  the  ruler  and  the  triangle,  construct  the  other  two 
sides. 

If  the  work  is  properly  done,  these  two  sides  will  measure 
2  in.  and  3  in.,  respectively;  and  the  remaining  angles  will 
measure  120°,  60°,  and  120°,  respectively.  From  this  exercise, 
the  pupils  should  learn  that  the  opposite  sides  and  the  opposite 
angles  of  a  parallelogram  are  equal,  and  that  the  sum  of  the 
four  angles  is  360°.  The  angles  being  oblique,  the  figure  is  a 
rhomboid. 

The  work  of  the  scholars  should  show  the  variety  suggested  in 
previous  exercises.     It  is  not  essential  that  the  longer  of  the  two 


NOTES  ON   CHAPTER  SIXTEEN  247 

given  sides  should  be  taken  as  the  base,  nor  that  the  base  should 
be  parallel  to  the  lower  edge  of  the  paper. 

57.  Different  pupils  will  construct  this  trapezoid  in  different 
ways.  Some,  seeing  that  one  angle  is  a  right  angle,  will  use  the 
triangle  to  draw  the  second  side,  3  in. ;  and  at  the  extremity  of 
this  side,  will  draw,  by  the  same  means,  an  indefinite  perpendic- 
ular line  to  form  the  third  side,  which  is  parallel  to  the  base. 
The  fourth  side  is  drawn  to  make  an  angle  of  60°  with  the  base. 

The  remaining  angles  will  measure  90°  and  120°,  respectively ; 
and  the  sides  will  measure  5  in.,  3  in.,  nearly  3J  in.,  and  nearly 
3iin. 

58.  The  triangle  cut  off  will  form  a  rhombus  when  opened  out, 
unless  the  base  and  the  perpendicular  are  equal.  In  this  case, 
the  paper,  when  opened,  will  form  a  square. 

Making  one  angle  of  the  triangle  30°  (or  40°)  will  give  a 
rhombus  containing  60°  (or  80°). 

60.  When  the  three  sides  of  a  triangle  are  equal,  its  three 
angles  are  equal ;  but  the  rhombus  has  four  equal  sides  without 
having  equal  angles. 

61.  A  triangle  that  contains  three  equal  angles,  has  its  sides 
equal ;  but  an  oblong  has  four  angles  of  90°  each,  with  unequal 
sides. 

62.  To  construct  the  rhomboid,  draw  a  base  of  2^  in.  Two 
inches  above,  draw  a  parallel  line  2-1-  in.  long,  with  the  extremi- 
ties of  the  latter  on  the  right  or  the  left  of  the  extremities  of  the 
base.  If  the  two  remaining  sides  are  exactly  equal  to  each 
other,  the  work  is  correctly  done. 

It  is  not  necessary  to  draw  the  altitude,  though  a  broken  line 
may  be  used.  While  different  methods  may  be  employed,  the  use 
of  an  incorrect  one  should  not  be  permitted.  The  work  done  on 
the  board  by  a  pupil,  should  be  criticised  by  the  class  if  it  be 
faulty ;  or  a  better  way  may  be  suggested,  if  the  one  employed  by 
the  pupil  at  the  board  requiie  too  much  time  or  unnecessary  work. 


248 


MANUAL   FOR   TEACHERS 


The  lengths  of  the  two  remaining  sides  should  be  measured, 
and  marked  on  the  papers.  Those  of  different  pupils  should  be 
different,  there  being  no  limit  except  the  size  of  the  paper: 
they  must,  however,  be  longer  than  2  in.  each ;  and  they  should 
not  be  just  2J  in.,  which  would  make  the  figure  a  rhombus. 

63.  If  the  line  that  shows  the  altitude  of  the  rhomboid  can 
be  drawn  within  the  figure,  a  right-angled  triangle  can  be  cut 
from  one  side  and  transferred  to  the  other,  making  the  figure  a 
rectangle.     Arithmetic,  Art.  929,  5,  last  parallelogram. 

In   the   case   of  a  rhomboid   whose    altitude    does    not   fall 


Fio.  6. 


within  the  figure,  several  cuts  will  be  necessary  to  change  it  into 
a  rectangle.     See  Fig.  5. 

65.  The  areas  will  be  equal,  because  each  rhomboid  is  equal  in 
area  to  a  rectangle  3  in.  by  2  in. 

66.  The  three  rhomboids  in  Fig.  6  have  their  respective  sides 
equal  each  to  each,  but  their  angles  are  unequal ;  hence  their 
altitudes  and  their  areas  are  unequal. 

To  construct  these  rhomboids,  draw  base  lines  of  three  inches, 


and  inclined  to  each,  at  any  angle  except  one  of  90°,  a  two-inch 
line.     Use  the  ruler  and  the  triangle  to  complete  the  figures. 


NOTES  ON   CHAPTER  SIXTEEN  249 

68.  As  the  altitudes  are  not  given,  the  areas  of  the  figures 
drawn  by  different  pupils  should  vary.  The  protractor  or  the 
triangle  should  be  used  in  drawing  the  altitudes,  which  should 
then  be  carefully  measured. 

69.  See  62,  making  the  upper  side  2^  in.  The  fourth  trape- 
zoid in  Arithmetic,  Art.  929,  6,  shows  the  manner  of  determining 
the  dimensions  of  the  required  rectangle. 

1267.  78.  The  diameters  of  these  circles  should  be  such  as 
not  to  admit  of  the  use  of  the  protractor  in  drawing  them.  To 
ascertain  the  extremities  of  an  arc  of  120°,  two  lines  are  drawn 
meeting  in  the  center  of  the  circle  at  an  angle  of  120°.  The  por- 
tion of  the  circumference  intercepted  by  these  lines  will  constitute 
an  arc  of  120°.  The  remainder  of  this  circumference  will  form 
an  arc  of  240°. 

79.  The  pupil  should  be  permitted  to  make  the  first  attempts 
in  his  own  way.  He  will  doubtless  soon  discover  that  the 
distance  between  the  points  of  his  compasses  in  drawing  the  cir- 
cle, is  the  length  of  the  chord  required,  and  that  by  placing  one 
point  of  the  compasses  on  any  portion  of  the  circumference  of  the 
circle  just  drawn,  the  other  point  will  indicate  on  the  circumfer- 
ence the  other  extremity  of  the  chord. 

To  measure  the  length  of  the  arc  in  degrees,  draw  radii  to  its 
extremities,  and  use  the  protractor  to  determine  the  angle  made 
by  these  radii,  which  may  be  produced,  if  necessary.  If  the  work 
is  properly  done,  the  angle  should  measure  60°, 
which  is  the  length  of  the  arc. 

80.  Draw  two  light  lines  meeting  at  an 
angle  of  72°.  Using  the  vertex  of  the  angle 
as  a  center,  and  any  radius,  draw  an  arc  be- 
tween the  lines.  This  arc  will  measure  72°. 
Darken  the  lines  from  the  arc  to  the  vertex 
to  show  the  radii  of  the  circle  of   which  the  arc  is  a  part. 


250 


MANUAL  FOR  TEACHERS 


81,  82.    Figs.  8  and  9  show  sectors  of  90°  and  270°,  respec- 
tively ;  Figs.  10  and  11,  segments  of  120°  and  240°,  respectively. 


Pig.  8. 


Fig.  9. 


Fig.  10. 


Fig.  11. 


1268.  86.  A  diameter  will  divide  the  circle  into  two  equal 
parts.  A  second  diameter  perpendicular  to  the  first,  drawn  by 
means  of  the  protractor  or  the  triangle,  will  divide  the  circle 
into  four  equal  parts. 

To  divide  the  circumference  into  four  equal  parts,  it  will  not 
be  necessary  to  draw  the  diameters.  When  he  has  the  ruler  in 
the  proper  place  to  draw  the  first  diameter,  the  pupil  needs  to 
mark  only  the  two  points  where  the  ruler  cuts  the  circumference. 
The  third  point  can  be  indicated  when  the  triangle  is  placed  at 
the  center  of  the  circle  and  against  the  ruler ;  etc. 

While  the  scholars  may  be  permitted  in  the  beginning  of  this 
work  to  draw  a  number  of  unnecessary  lines,  and  while  it  may 
be  an  advantage  to  even  require  it,  they  should  gradually  learn 
to  make  as  few  lines  as  possible.  The  construction  lines  that 
are  employed,  should  be  drawn  very  lightly  and  should  not  be 
erased.  Other  lines  should  be  made  more  conspicuous.  Careful 
pupils  may  be  allowed  to  use  ink  for  this  purpose. 

1269.  It  is  not  intended  that  the  methods  here  suggested 
should  be  communicated  in  advance  to  pupils.  Each  should  be 
allowed  to  try  the  problem  in  his  own  way.  The  discussion  of 
the  method  employed  afterwards  on  the  blackboard,  will  suggest 
other  and  possibly  better  modes  of  procedure. 

88.  To  inscribe  a  regular  pentagon  in  a  circle,  it  will  be  neces- 
sary to  divide  the  circumference  into  five  arcs  of  72°  each.     The 


NOTES  ON   CHAPTER  SIXTEEN  251 

protractor  should  be  used  to  obtain  the  first  arc ;  the  remain- 
ing ones  can  be  set  off  by  the  compasses,  the  first  being  used  as 
a  measure.     Careful  work  should  be  exacted  by  the  teacher. 

89.  Many  pupils  will  have  learned  in  their  drawing  lessons 
the  regular  method  of  inscribing  a  hexagon  in  a  circle.  Those 
unfamiliar  with  this  way,  should  not  be  shown  it  until  after  they 
have  constructed  the  hexagon  by  means  of  the  protractor. 

It  is  a  pedagogical  mistake  to  suggest  "  short-cuts  "  to  pupils 
before  they  thoroughly  understand  a  general  method.  For  this 
reason,  the  teacher  should  permit  the  members  of  her  class  to  use 
the  protractor  in  the  construction  of  the  inscribed  triangle,  leav- 
ing it  to  themselves  to  discover  a  simpler  way.  She  should 
encourage,  also,  the  employment  of  a  variety  of  methods  even 
if  some  of  them  are  not  very  direct.  The  experiments  made  by 
pupils  to  discover  a  new  mode  of  constructing  a  polygon,  will 
help  them  in  their  geometrical  study. 

The  chord  of  60°  being  equal  in  length  to  the  radius  (79), 
the  shortest  method  of  inscribing  a  hexagon  is  to  apply  the 
radius  as  a  chord  six  times.  Two  of  these  divisions  of  the  circum- 
ference will  make  arcs  of  120°,  the  chords  of  three  of  these  form- 
ing the  sides  of  an  inscribed  equilateral  triangle, 

90.  Each  of  the  six  angles  at  the  center  contains  60°.  Since 
the  two  sides  AC  and  AB,  enclosing  any  central  angle,  are  radii 
of  the  circle,  and  therefore  equal  to  each  other,  the  angles  oppo- 


FiG.  12.  Fig.  13. 


site  those  sides  are  equal ;  that  is,  angle  A  =  angle  B.  The 
angle  at  C  being  60°,  A-\-B  =  180°  -  60°  =  120°,  and  ^  =  ^  - 
60°.     Angles  1  and  2  (see  Arithmetic),  therefore,  measure  60° 


252  MANUAL   FOR   TEACHERS 

each,  and  the  whole  angle  contained  between  two  adjoining 
sides  of  the  hexagon,  measures  120°.  After  determining  by  this 
method  the  number  of  degrees  in  each  angle  of  a  regular  hexa- 
gon, the  pupils  should  be  required  to  construct  one,  and  to  mark 
in  each  angle  its  contents  in  degrees,  as  in  Fig.  13,  verifying  the 
result  by  using  the  protractor. 

91.  A  careless  scholar,  measuring  the  number  of  degrees  in 
each  angle  of  a  regular  pentagon  (Arithmetic,  Art.  1268),  will 
sometimes  read  from  the  wrong  row  of  numbers  on  the  protrac- 
tor, getting  the  result  72°,  instead  of  the  correct  one  of  108°. 
As  there  are  72°  in  each  division  of  the  circumscribing  circle,  he 
will  have  no  doubt  of  the  correctness  of  his  answer,  unless  he 
has  been  trained  to  estimate  the  size  of  an  angle.  In  this  case, 
he  will  see  that  each  angle  of  a  regular  pentagon  is  obtuse,  and, 
therefore,  greater  than  90°. 

One  method  of  calculating  the  number  of 
degrees,  is  to  divide  the  pentagon  into  five 
equal  triangles,  one  of  which  is  shown  in  Fig. 
14.  The  angle  at  Cis  72°.  The  sides  CA  and 
CB,  being  radii,  are  equal ;  which  makes  equal  ^^'  ^*" 

angles  at  A  and  B,  each  of  which  is  ^  of  (180°  -  72°),  or  54°. 
Each  of  these  angles  is  the  half  of  one  of  the  angles  of  the  pen- 
tagon, so  that  these  latter  angles  measure  108°  each. 

92.  A  circumscribed  square  touches  the  circle  at  four  points, 
each  side  constituting  a  tangent.  A  tangent  being  perpendicular 
to  the  radius  drawn  to  the  point  of  contact,  the  square  may  be 
constructed  by  drawing  perpendiculars  to  two  diameters  inter- 
secting at  right  angles,  using  the  triangle  or  the  protractor  for 
the  purpose.  The  ingenious  pupil  will  discover  other  ways ; 
drawing,  for  instance,  at  each  extremity  of  the  two  diameters,  a 
line  parallel  to  the  intersecting  diameter,  by  means  of  the 
ruler  and  the  triangle;  etc.  No  method  should  be  permit- 
ted that  merely  approximates  accuracy,  such  as  determining 
that  a  line  is  parallel  or  perpendicular  by  the  eye  alone.     The 


NOTES   ON   CHAPTER   SIXTEEN 


253 


Fig.  15. 


Through  3  and 


average  class  will  contain  many  members  intelligent  enough  to 
pass  upon  the  correctness  of  a  given  method,  and  they  should  be 
called  upon  to  give  reasons  for  any  criticisms  they  may  have 
to  offer. 

In  circumscribing  some  polygons,  a  hex- 
agon for  instance,  many  pupils  prefer  locat- 
ing points  X  and  Fj  instead  of  using  the 
triangle  and  the  ruler  to  draw  a  tangent 
XF.  After  dividing  the  circumference 
(Fig.  15)  into  six  equal  parts  at  1,  2,  3,  4,  5, 
and  6,  they  draw  a  diameter  from  1  to  4. 
Through  2  and  6  they  draw  a  secant  XA, 
making  MX  and  MA  each  equal  to  the  radius. 
5  a  second  secant  is  drawn,  and  NY  and  NB  are  also  made 
equal  to  the  radius.  A  line  drawn  through  X  and  Y  will  form 
one  side  of  the  circumscribed  hexagon.  One  extremity  of  this 
side  can  be  determined  by  producing  the  diameter  3(76  to  F, 
and  the  other  by  a  line  through  2(75. 

Note.  —  A  secant  is  a  line  that  cuts  a  circle  at  two  points ;  a  tangent  is 
a  line  that  touches  a  circle  at  one  point. 

93.  The  smallest  number  of  triangles  into  which  a  pentagon 
can  be  divided,  is  three  (Fig.  16).  The 
three  angles  of  each  triangle  contain  180°, 
making  the  sum  of  angles  1-9,  540°. 
Since  1  =  A,  2  +  4:  =  B,  6  +  7  =  C,  8  = 
I);  3 -f  5  =  jE",  the  sum  of  the  five  equal 
angles  (A,  B,  C,  D,  and  ^)  of  a  regular 
pentagon  =  540°,  and  each  equals  108°. 
This  is  the  result  that  was  found  in  91 
by  another  method. 

94.  The  hexagon  is  divisible,  as  above,  into  four  triangles, 
containing  180°  X  4,  or  720° ;  making  each  angle  720°  h-  6, 
or  120°. 

95.  A  quadrilateral  is  divisible  into  2  triangles ;  a  pentagon, 


254 


MANUAL   FOR   TEACHERS 


into  3 ;  a  hexagon,  into  4 ;  a  heptagon,  into'  5 ;  an  octagon,  into 
6,  beiiig  2  triangles  less  in  each  case  than  the  number  of  sides  in 
the  polygon. 

96.  The  number  of  degrees  in  each  angle  of  a  regular  octagon 
=  [180°  X  (8 -2)]  ^8. 

97.  At  each  end  of  the  2-inch  line,  draw  a  2-inch  line  at  an 
angle  of  108°.  At  the  farther  extremity  of  each  of  those  lines 
draw  a  line  at  an  angle  of  108°.  These  last  lines  meet  at  an 
angle  of  108°  if  the  work  is  correctly  done,  and  are  each  two 
inches  long  to  the  point  of  their  intersection. 

98.  A  line  drawn  to  each  extremity  of  the  base  at  an  angle 
of  60°  will  form  an  equilateral  triangle.  Use  angles  of  90°  for 
the  square,  108°  for  the  pentagon,  120°  for  the  hexagon,  nearly 
129°  for  the  heptagon,  135°  for  the  octagon,  140°  for  the  non- 
agon,  etc. 

The  first  line  should  be  placed  near  the  cen- 
ter of  the  bottom  of  the  paper  to  give  room  for 
the  successive  polygons.  See  Fig.  17.  Bright 
pupils  will  calculate  the  necessary  angles  and 
continue  to  construct  polygons  as  far  as  the 
space  will  permit. 

99.  By  drawing  the  diameters  AB  and  XY  (Fig.  18),  the 
inscribed  square  will  be  divided  into  four 
triangles,  while    the    circumscribed  square 
contains  eight,  being  double  the  area  of  the 
inscribed  square. 

1270.  These  problems  are  given  to  en- 
able the  pupils  to  learn  how  to  bisect  lines, 
erect  perpendiculars,  construct  angles,  etc., 
by  means  of  the  ruler  and  the  compasses, 
and  incidentally  to  learn  a  number  of  geometrical  facte.  The  use 
of  other  instruments  is  unnecessary,  and  should,  therefore,  not 
be  tolerated. 


Fig.  17. 


NOTES   ON   CHAPTER   SIXTEEN  255 

1.  The  distance  between  the  centers  ==  1^  in.  +  1  in. 

2.  11  in.  —  1  in. 

4.  The  line  XY  joining  the  two  points  of  intersection  of  the 
equal    circles  (Fig.    19),  bisects   the 
line    AB    connecting    the    centers. 
The  radii  AX,   BX,  A  Y,  BY  are 
each  2  inches  long. 

5.  The  previous  exercise  should 
lead  the  pupils  to  see  the  steps  neces- 
sary to  the  construction  of  the  re- 
quired triangle.  The  3-inch  base  AB  is  first  drawn.  The  next 
requirement  is  to  find  a  point  X(Fig.  19)  2  inches  from  A  and 
from  B.  A  circle  of  2-inches  radius  with  i>  as  a  center,  will 
contain  every  point  that  is  2  inches  from  B.  A  similar  circle 
with  ^  as  a  center,  contains  every  point  2  inches  from  A. 
The  intersections,  X  and  Y,  are  each  2  inches  from  both 
points  A  and  B.  Using  either  one  as  a  vertex,  draw  lines  to 
A  and  B,  forming  the  required  triangle. 

Authorities  differ  somewhat  as  to  the  advisability  of  requiring  pupils  to 
employ  circles  rather  than  arcs  in  geometrical  problems.  While  it  may  be 
better  at  first  to  use  circles,  the  point  to  be  finally  reached  is  the  emj^loy- 
ment  of  the  shortest  lines  possible.  This  should  not  be  inconsistent  with 
the  acquirement  of  geometrical  knowledge.  A  scholar  should  know  after 
a  very  few  exercises  that  each  point  of  an  arc  is  2  inches  from  the  center, 
as  well  as  he  does  when  he  draws  the  entire  circle. 

In  his  later  construction  of  an  isosceles  triangle,  the  pupil  should  know 
that  the  vertex  is  above  (or  below)  the  center  of  the  base.  For  this  reason 
the  first  arc  need  not  be  a  very  long  one.     The  second  should  be  still  shorter. 

6.  Fig.  19  will  suggest  the  necessary  steps.  Using  each  end 
of  the  3^-incli  base  as  a  center,  and  with  a  radius  of  4  inches, 
draw  two  circles.  Draw  lines  corresponding  to  XA  and  XB  for 
the  required  triangle.  Placing  the  ruler  on  X  and  Y  will  give 
the  perpendicular,  which  should  not  extend  below  the  base. 

If  arcs  are  used,  the  upper  intersection  determines  the  position 
of  the  vertex.     A  lower  intersection  is  used  to  determine  the 


256  MANUAL   FOR  TEACHERS 

direction  of  the  perpendicular.  The  pupil  will  gradually  learn 
that  while  a  definite  radius,  4  inches,  is  required  to  locate  the 
vertex  of  the  triangle,  intersecting  arcs,  each  of  3  inches  or  5 
inches  or  any  other  radius,  will  serve  to  locate  the  second  point, 
used  with  the  vertex  to  determine  the  direction  of  the  perpendic- 
ular. The  point  of  their  intersection  may  be  above  the  base 
or  below  it,  according  to  convenience.  A  point  below  secures 
greater  accuracy,  by  being  probably  at  a  greater  distance  from 
the  vertex  than  one  above  is  likely  to  be. 

7.  This  is  a  variation  of  6,  but  without  directions  as  to  length 
of  sides.  If  the  perpendicular  is  correctly  drawn,  it  will  bisect 
the  base.  See  Exercise  48,  Art.  1263.  The  compasses  should 
be  employed  to  determine  the  equality  or  inequality  of  the  seg- 
ments of  the  base. 

8.  The  same  procedure  is  required  as  in  7,  except  that  the 
sides  of  the  triangle  are  not  drawn.  The  bisecting  line  should 
be  extremely  short. 

9.  With  a  2-inch  radius,  draw  intersecting  arcs,  using  as 
centers  the  two  extremities  of  a  2-inch  line. 

11.  Either  side  may  be  used  as  the  base  ;  and  different  pupils 
should  use  a  different  base,  although  the  greatest  number  will 
probably  take  the  longest  side. 

Using  the  2-inch  side  as  a  base,  draw  from  one  end,  as  a  center, 
an  arc  with  a  radius  of  1  inch ;  and  from  the  other,  an  arc  with 
a  radius  of  1 J  inches.  The  intersection  of  the  arcs  will  be  the 
vertex  of  the  required  triangle. 

With  the  3-inch  line  as  a  base,  the  radius  of  the  arcs  will  be 
2  inches  and  2-J  inches,  respectively. 

Besides  employing  different  bases,  the  pupils  should  use  oblique  lines  and 
vertical  lines  as  bases,  and  the  vertex  in  some  instances  should  be  below  the 
base ;  etc. 

12.  A  scholar  should  be  permitted  to  discover  for  himself  that 
the  intersection  of  the  2-inch  arcs  will  be  at  the  center  of  the 
4-inch  base,     After  endeavoring,  also,  to  make  a  triangle  whose 


NOTES   ON   CHAPTER   SIXTEEN  257 

sides  shall  measure  1,  2,  and  3  inches,  respectively,  he  will 
understand  that  the  third  side  of  a  triangle  must  be  shorter  than 
the  combined  lengths  of  the  other  two. 

13.  If  the  pupil,  in  drawing  arcs  to  locate  the  bisecting  line, 
employs  the  radius  used  in  drawing  the  circle,  he  will  discovGr 
that  one  intersection  will  take  place  at  the  center  of  the  circle. 
This  will  lead  him  to  see  that  only  one  set  of  intersecting  arcs  is 
necessary  —  the  one  beyond  the  circle,  the  center  of  the  circle 
serving  as  a  second  point. 

The  teacher  should  be  in  no  hurry  to  inform  the  pupil  using 
two  sets  of  arcs,  that  a  single  set  will  be  sufficient ;  knowledge 
that  comes  in  some  other  way  than  merely  by  direct  telling,  is 
apt  to  last  longer. 

14.  Before  dividing  a  sector  (Figs.  8  and  9)  into  two  equal 
parts,  some  scholars  may  consider  it  necessary  to  draw  the  chord. 
Permit  them  to  do  so  at  first. 

15.  The  pupils  have  already  learned  that  a  line  drawn  from 
the  vertex  of  an  isosceles  triangle  to  the  center  of  the  base,  is  per- 
pendicular to  the  base.  The  method  employed  in  bisecting  a 
line  is  practically  to  consider  it  the  common  base  of  two  isosceles 
triangles  having  their  vertices  on  opposite  sides  of  the  base, 
although  the  equal  sides  of  the  triangles  are  not  drawn.  This 
bisecting  line  is  perpendicular  to  the  assumed  base.  The  pupils 
have  probably  discovered  that  the  perpendicular  line  bisecting 
the  chord  in  13,  would,  if  produced,  pass  through  q 

the  center  of  the  circle.     From   this  previous 
knowledge,  they  will  doubtless  be  able  to  answer  / 

the  last  question  of  15.  / 

The  scholars  that  have  drawn  the  chord  in     \II___ 
dividing  the  sector  in  14  into  two  equal  parts,  ^ 

will  be  likely  to  see  that  the  radius  CX  (Fig.  fig.  20. 

20)  is  perpendicular  to  the  bisected  chord  AB. 

16  shows  that  a  perpendicular  that  bisects  a  chord,  AB  (Fig. 
20),  also  bisects  the  arc  AJCB. 


258  MANUAL   FOR   TEACHERS 

It  will  readily  be  seen  by  the  pupils  that  an  arc  AXB  can  be 
bisected  without  drawing  the  chord  AB. 

17.  This  problem  is  the  same  as  8.  The  drawing,  however, 
should  show  a  longer  line,  and  one  that  does  not  cut  the  given 
line,  the  requirement  being  that  the  perpendicular  be  drawn  io 
the  latter.  A  perpendicular  drawn  to  a  horizontal  or  to  an 
oblique  line  from  below  it,  would  be  correct ;  although  it  might 
not  be  accepted  as  satisfactory  if  the  more  common  wording  of 
the  problem  were  employed :  Erect  a  perpendicular  at  the  mid- 
dle point  of  a  line. 

18  requires  no  explanation. 

19.  Bisecting  one  of  the  four  divisions  of  a  circumference 
gives  the  dividing  point  between  two  one-eighths.  A  ruler  placed 
upon  this  point  and  on  the  center  of  the  circle  will  indicate  the 
location  of  another.  The  distance  between  two  points  can  be 
ascertained  by  the  compasses,  which  can  then  be  used  to  locate 
the  remaining  two  dividing  points. 

The  lines  used  to  divide  a  circumference  should  not  be  too  long. 

The  thoughtless  pupil  sometimes  fails  to  see,  when  he  employs 
his  compasses  to  measure  the  distance  between  two  points  on  a 
circumference,  that  he  is  measuring  the  chord  of  an  arc,  and  not 
the  arc  itself.  It  may  be  necessary  for  tJie  teacher  to  explain 
this  in  connection  with  25. 

22.  Unless  the  pupils  have  learned  in  their  drawing  work  the 
method  of  erecting  a  perpendicular  at  the  end  of  a  line,  some  of 
them  may  experience  a  little  difficulty  in  solving  this  problem. 
One  way  of  drawing  the  circumscribed  square  is  to  construct 
on  AY  (Fig.  18)  an  isosceles  triangle  A2Y  equal  to  ACY. 
Produce  2 A  to  1,  making  -41  equal  to  2 A.  Produce  2Y  to 
3  in  the  same  manner.  Lines  from  1  through  JT,  and  3  through 
B,  will  intersect  at  4,  which  completes  the  square. 

24.  The  object  of  23  and  24  is  to  lead  the  pupils  to  see  again 
that  the  side  of  an  inscribed  hexagon  is  equal  in  length  to  the 
radius  of  the  circle. 


NOTES   ON   CHAPTEK   SIXTEEN  259 

26.  The  four  triangles  will  together  constitute  a  circumscribed 
equilateral  triangle. 

27.  Draw  lightly  an  arc  less  than  90°  in  length.     After  cut- 
ting off  60°,  darken  this  portion.     No  radii   ^ 
should  be  drawn. 

To  construct  an  angle  of  60°,  draw  an  in- 
definite line  AB  (Fig.  21).  With  any  con- 
venient radius,  as  AM,  draw  an  arc  Mli. 
With  Jf  as  a  center,  and  using  the  same 
radius,  cut  the  arc  at  X.  This  makes  3fX  an  arc  of  60°.  From 
A  draw  a  line  through  X. 

28.  An  arc  of  30°  is  constructed  by  the  bisection  of  an  arc 
of  60°. 

To  construct  an  angle  of  30°,  draw  AB  (Fig.  21)  and  the  arc 
MN;  and  cut  off  MX,  as  in  27.  With  the  same,  or  any  other 
convenient  radius,  and  using  M  and  X  as  centers,  draw  two 
arcs  intersecting  on  the  right  of  MX.  A  line  drawn  from  A 
through  this  intersecting  point  will  make  with  AB  an  angle 
of  30°. 

29.  To  construct  an  angle  equal  to  60°  +  30°  (or  30°  +  60°), 
bisect  the  arc  XM  (Fig.  21),  and  from  the  bisecting  point,  which 
is  30°  from  M,  lay  off  an  additional  60°.  A  line  drawn  from  A 
through  this  point  will  make  with  AB  an  angle  of  90°. 

30.  One  method  of  erecting  a  perpendicular  at  the  end  of  a 
line    is   given    in    29.      A   somewhat 

similar  method  consists  in  prolonging 
the  arc,  and  marking  off  at  Q  and 
B  two  divisions  of  60°  each.  Using 
these  two  points  as  centers,  draw  arcs 
intersecting  at  F.  From  I)  draw  a 
line  through  i^. 

The  line  BF  bisects  the  arc  QB, 
which  makes  the  angle  FDE  = 
60°(PQ)+30°(QF)-90°. 


260  MANUAL   FOR   TEACHERS 

31.  The  bisection  of  VF  gives  an  angle  of  45°.  Do  not 
draw  DF. 

The  recommendation  so  often  made  as  to  getting  a  variety  of  drawings 
from  the  various  members  of  the  class,  applies  to  these  problems.  The 
erection  of  a  perpendicular  at  the  end  of  a  vertical  line  should  call  forth 
at  least  four  different  results.  The  perpendiculars  may  be  erected  at  either 
end,  and  may  run  to  the  right  or  to  the  left.  In  constructing  an  angle  of 
45°,  a  greater  variety  is  possible.  The  first  line  may  be  horizontal,  verti- 
cal, or  oblique ;  the  second  line  may  start  from  either  end ;  and  it  may 
extend  above  or  below,  to  the  right  or  to  the  left;  the  lines  forming  the 
angles  need  not  be  of  the  same  length  on  all  papers,  nor  need  both  of  them 
be  of  the  same  length  on  any  one  paper. 

22^°  =  I  of  45°;  135°  =  90°  +  45° ;  15°  =  |  of  30°;  75°  = 
60° +15°. 

33.  The  preceding  problem  should  give  a  hint  as  to  the 
method  of  solving  the  present   one.     In  ^ 

32,  a  circle  was  drawn  with  ^  as  a  center 
and   AX  as   a   radius.     This    circle   cut 
XY  (Fig.  23)  in  M.     To  erect  a  perpen- 
dicular at  A,  the  center  of  the  circle,  the  ^'  a         M         ' 
extremities   X  and    M  of  the   diameter  ^^'  ^" 
XM  were  used  as  centers  to  draw  arcs  intersecting  at  J.     A 
perpendicular  was  then  drawn  from  A  through  J. 

This  problem  does  not  furnish  a  circle,  nor  is  one  necessary. 
The  lino  XY  is  given,  and  the  point  A  at  which  the  perpendic- 
ular is  to  be  erected.  With  ^  as  a  center,  and  a  radius  equal 
to  AX,  cut  off  AM.  This  gives  us  the  diameter  of  the  circle 
employed  in  the  preceding  problem. 

It  will  be  noticed  that  a  second  set  of  intersecting  arcs  on  the 
opposite  side  of  XM  is  not  needed,  the  point  A  answering 
instead.  This  problem  amounts  to  the  bisection  of  an  arc  of 
180°,  of  which  only  the  chord  XM  is  drawn,  A  being  the  center 
of  the  circle. 


NOTES  ON  CHAPTER  SIXTEEN 


261 


34.  The  first  two  divisions  of  this  problem  are  reviews  of 
parts  of  17  and  30.     The  erection  j^ 

of    a   perpendicular    at    a   point 

between  the  end  and  the  center, 

is   a  variation    of  the   preceding 

problem.     Let   BS  (Fig.    24)  be 

the    required     line    and    A    the 

point.     With  ^  as  a  center,  and 

any  convenient  radius,  lay  off  the  Fig.  24.  '^ 

points  Jl   and  M,  which  will  be 

equidistant  from   A.     Using   X  and   M  as   centers,  draw  arcs 

intersecting  at  tT.     Draw  the  required   perpendicular  from   A 

through  J. 

In  problem  33  the  point  M  was  located  at  a  distance  from  A,  equal  to 
AB\  and  while  greater  accuracy  is  obtained  by  having  the  points  J/and 
X  as  far  apart  as  possible,  the  present  method  is  suggested  to  show  that  the 
only  essential  requirement  as  to  their  location  is  to  have  them  equidistant 
from  A. 

35.  The  base  need  not  be  a  horizontal  line. 

36.  Proceed  as  in  the  construction  of  a  right-angled  triangle, 
base  2  inches,  perpendicular  2  inches  ;  but  do  not  B 
draw  the  hypotenuse.     With  B  and  H  as  cen- 
ters (Fig.  25),  and  a  radius  of  2  inches  (BP  or 
BIT),  draw  arcs  intersecting  at  0.     BO  and  ITO 
will  form  the  remaining  sides. 

To   construct   the     rectangle,    BH  (Fig.  25) 
should  be  made  3  inches  long,  BB  measuring  2  Fig.  25. 

inches.  With  ^  as  a  center,  and  a  radius  of  3  inches,  draw  an 
arc.  Intersect  this  by  a  second  arc  drawn  with  IT  as  a  center, 
and  a  radius  of  2  inches.  From  the  point  of  intersection,  draw 
lines  to  B  and  IT. 

Some  scholars  may  prefer  to  erect  a  perpendicular  at  each  end  of  the 
base,  etc. 

38.  Draw  a  base  line,  WV  (Fig.  26),  3  inches  long.  At  any 
convenient  point,  M,  erect  a  perpendicular,  MB,  2|-  inches  long. 


P 


H 


262 


MANUAL   FOR  TEACHERS 


At  its  extremity  P,  draw  the  perpendicular  PO.  With  TT  as  a 
center  and  a  radius  of  3  inches,  locate  the  point  Xon  the  line 
P0\  TP'Xwill  be  the  second  side  of  the  rhombus.  On  PO  draw 
XN  3  inches;  and  connect  NV.  This  last  line  should  measure 
3  inches  if  the  work  is  properly  done. 


Fig. 

To  construct  a  3-inch  rhombus  containing  an  angle  of  60°, 
draw  WV\  at  W  construct  an  angle  of  60°,  and  make  WX 
3  inches.  At  X  or  at  F",  draw  a  3-inch  line  making  an  angle  of 
120°  with  XW  OY  VW,  etc. ;  or,  using  Xand  Fas  centers,  and 
with  a  radius  of  3  inches,  draw  arcs  intersecting  at  iV;  draw 
iVXandATF. 

39.  Proceed  as  in  the  first  problems  of  38  ;  but  make  TTFand 
XN  each  4  inches  long. 

40.  See  PPB  (Fig.  25)  ;  draw  PIL 

Some  pupils,  preferring  to  commence  with  a  horizontal  base, 
calculate  the  number  of  degrees  in  each  angle  at  the  base, 
i  (180°  -  90°),  or  45°.  To  each  end  of  a  base  line  of  any  length 
(Fig.  27),  they  draw  a  line  making  with  the 
base  an  angle  of  45°. 

If  one  angle  is  120°,  the  angles  at  the  base 
will  be  30°  each;  if  one  is  135°,  the  angles  at 
the  base  will  measure  22^°  each. 

The  method  shown  in  Fig.  27  requires  the  construction  of  two 
angles ;  and  is,  therefore,  not  so  direct  as  drawing  two  equal  lines 
meeting  at  the  given  angle. 


Fio.  27. 


NOTES   ON   CHAPTER  SIXTEEN 


263 


B  Y 

Fig.  28. 


41.  Erect  a  3-inch  perpendicular  at  the  middle  of  a  3-inch 
line.  A 

A  perpendicular,  AB  (Fig.  28), 
bisects  the  angle  at  A  of  the  equilat- 
eral triangle.  Erect  a  3-inch  perpen- 
dicular at  By  on  an  indefinite  line  ^ 
CD;  at  A,  construct  two  angles  of 
30°  each,  as  shown  in  the  figure ;  draw  AX  and  A  Y,  forming 
the  required  equilateral  triangle  AXY.  Test  the  work  by  meas- 
uring XY. 

42.  To  construct  a  scalene  triangle  of  the  required  dimen- 
sions, erect  the  3-inch  perpendicular  at  a  point  not  in  the  center 
of  the  3-inch  base,  nor  at  either  extremity. 

To  construct  the  obtuse-angled  triangle,  produce  the  3-inch 
base  by  a  dotted  line,  and  erect  the  3-inch  altitude  at  some  point 
outside  of  the  base. 


43.  This  problem  may  puzzle  the  pupils  at  first.  If  the  tri- 
angle were  isosceles,  XY3f,  for 
instance,  there  would  be  no  diffi- 
culty. The  solution  of  the  prob- 
lem consists  practically  in  making 
an  isosceles  triangle,  although  the 
line  X3f\s  not  drawn.  With  X 
as  a  center  (Fig.  29),  and  XYas 
a  radius,  the  point  M  is  located 
(without  drawing  the  arc  shown 
in  the  figure).  Using  If  and  Y  as  centers,  and  with  any  con- 
venient radius,  arcs  are  drawn  intersecting  at  B.  XB  gives  the 
direction  of  the  altitude. 


Fig.  29. 


44.  The  given  triangle  in  this  case  will  be  XMZ  (Fig.  29). 
Produce  ZM  indefinitely  towards  Y,  and  with  XM  as  a  radius, 
cut  the  base  at  Y;  etc.  XFis,  of  course,  not  drawn  ;  il[fF  should 
be  a  light  line  or  a  broken  line ;  etc. 


264  MANUAL   FOR  TEACHERS 

45.  This  is  the  same  problem  in  another  form.  Let  YZ 
(Fig.  29)  be  the  given  line,  and  X  the  given  point,  from  which  a 
perpendicular  is  to  be  let  fall  upon  YZ.  In  43-44,  the  arc 
described  with  X  as  a  center,  passes  through  one  extremity  of 
the  line.  The  pupils  should  see  that  this  is  not  essential,  and 
that  it  would  be  very  inconvenient  in  the  case  of  a  very  long 
line.  All  that  is  necessary  is  to  intersect  the  line  YZ  at  two 
points  sufficiently  far  apart  to  secure  accuracy. 

47.  Each  angle  will  measure  60°;  the  arc  on  which  it  stands 
measures  120° ;  an  angle  at  the  circumference,  therefore,  is 
measured  by  one-half  the  arc  intercepted  by  its  sides. 

Apply  this  to  the  regular  pentagon  (Arithmetic,  Art.  1268). 
The  angle  at  the  top  of  the  figure  stands  on  an  arc  containing 
72°  +  72°  +  72°,  or  216°  (made  up  of  three  arcs)  ;  it  contains, 
therefore,  \  of  216°,  or  108°.     See  Fig.  30. 

An  angle  of  a  square  stands  on  an  arc  of  180° ;  its  contents  are, 
therefore,  \  of  180°,  or  90°  ;  etc. 


Fio.  80.  Fig.  81. 

48.  The  number  of  degrees  in  each  angle  of  a  regular  hexagon 
is  120°  (94,  Art.  1269).  From  each  end  of  the  given  line  AB 
(Fig.  31)  draw  a  line  equal  in  length  to  AB,  and  making  with  it 
an  angle  of  120°  ;  etc. 

While  this  is  not  the  best  way  to  construct  a  regular  hexagon, 
it  gives  the  pupils  an  opportunity  to  try  the  general  method  on  a 
polygon  of  six  sides. 

49.  For  the  construction  of  an  octagon,  the  angles  at  A,  B, 
C,  etc.,  should  measure  135°  (Problem  31). 


NOTES   ON   CHAPTER   SIXTEEN  2G5 

By  a  later  problem,  the  pupils  will  learn  how  to  find  the  center  of  the 
circumscribing  circle  after  two  sides  of  the  octagon  are  drawn  ;  for  the 
present,  however,  they  should  be  required  to  repeat  the  construction  of 
the  angle  of  135°  at  the  required  number  of  corners.  The  work  should  be 
as  accurate  as  possible,  considering  the  necessary  imperfections  of  the  tools 
employed. 

50.  A  right  angle  whose  vertex  is  at  the  circumference,  sub- 
tends an  arc  of  180°,  or  a  semi-circumference. 

51.  By  means  of  a  ruler,  placed  on  the  center  of  the  circle, 
mark  points  A  and  B  (Fig.  32)  the  extremi- 
ties of  a  diameter  (not  drawn).  The  arc 
A  MB  is  one-half  of  the  circumference,  or 
180° ;  and  the  angles  X,  Y,  and  Z,  whose 
sides  XA  and  XB,  YA  and  YB,  ZA  and 
ZB,  subtend  this  arc,  are  each  measured  by 
one-half  of  it,  and  contain,  therefore,  \  of 
180°,  or  90°. 

52.  The  hypotenuse  of  an  inscribed  right- 
angled  triangle  is  always  the  diameter  of  the  circle. 

In  Fig.  32  draw  the  diameter  AOB  (or  any  other).  Bisect 
the  arc  AXYZB,  which  gives  the  vertex  of  the  required  trian- 
gle.    Connect  this  point  with  the  points  A  and  B. 

Note.  —  It  may  be  necessary  to  have  the  scholars  understand  that  all 
the  vertices  of  an  inscribed  polygon  must  lie  in  the  circumference ;  an 
inscribed  triangle,  therefore,  has  three  vertices  in  the  circumference. 

53.  The  diameter  of  a  circle  having  a  radius  of  2  inches,  will 
bu  the  diagonal  of  the  inscribed  square.  The  location  of  the 
remaining  two  corners  is  easily  determined. 

The  square  may  also  be  constructed  by  drawing  4-inch  diagonals 
bisecting  each  other  at  right  angles  ;  etc. 

The  rhombus  will  consist  of  two  8-inch  equilateral  triangles 
having  a  common  base. 

54.  Every  angle  being  on  the  circumference,  each  will  be 
measured  by  one-half  the  intercepted  arc,  and  the  sum  of  the 


266  MANUAL   FOR  TEACHERS 

three  angles  will  contain  one-half  the  number  of  the  degrees  con- 
tained in  the  three  arcs  ;  i.e.,  ^  of  360°. 

65.    See  the  rhombus  in  53. 

56.  The  two  triangles  will  be  equal  in  all  respects,  as  will 
those  in  57. 

58.  From  56  and  57  the  pupils  should  learn  that  two  triangles 
are  equal  in  every  respect  (a)  when  two  sides  and  the  included 
angle  of  one  are  equal  to  two  sides  and  the  included  angle  of  the 
other,  each  to  each ;  and  (b)  when  two  angles  and  the  included 
side  of  one  are  equal  to  two  angles  and  the  included  side  of  the 
other,  each  to  each.  From  59,  they  should  learn  that  when  two 
triangles  have  the  three  sides  of  the  one  equal  to  three  sides 
of  the  other,  each  to  each,  they  are  equal  throughout.  From  the 
present  problem,  they  should  learn  that  any  number  of  triangles 
can  be  constructed  having  their  corresponding  angles,  equal  each 
to  each,  but  whose  corresponding  sides  are  unequal,  each  to  each. 

60.  Impossible,  because  the  sum  of  three  angles  of  75°  each  is 
greater  than  180°. 

61.  The  radius  of  each  circle  must  be  one-half  of  the  hypote- 
nuse, or  1-J-  in.     See  Problem  52. 

Any  inscribed  triangle  having  one  side  the  diameter  of  the 
circle  will  answer  the  conditions  of  the  first  case. 

The  2-inch  base  of  the  second  triangle  is  obtained  by  taking  A 
as  a  center  (Fig.  32),  and  with  a  radius  of  2  inches,  drawing 
an  arc  at,  say,  F;  ^Fwill  be  the  required  base,  Y£  the  per- 
pendicular, and  AB  the  hypotenuse. 

Using  a  radius  of  1^  inches,  as  above,  will  give  a  perpendic- 
ular, AX,  of  1^  inches,  etc. 

Note. — The  term  base  does  not  necessarily  imply  a  horizontal  line;  nor 
perpendicular,  a  vertical  one. 

63.  To  construct  an  angle  at  F(Fig.  33)  equal  to  the  angle 
at  X,  take  Xas  a  center  and  any  convenient  radius,  and  draw 
the  arc  Aa ;  then  take  F  as  a  center  and  the  same  radius,  and 
draw  the  arc  Gc.     The  angle  X  is  measured  by  the  arc  AB,  the 


NOTES   ON   CHAPTER   SIXTEEN 


267 


length  of  whose  chord  is  determined  by  means  of  the  compasses. 
Lay  off  the  same  on  Cb,  making  the  arc  CD  equal  to  the  arc 
AB.     An  angle  formed  by  drawing  a  line  from  Y  through  D 


Fio.  33. 

will  be  equal  to  the  angle  at  JT,  as  each  angle  is  measured  by  an 
arc  of  the  same  number  of  degrees.  See  note  to  Problem  19, 
Art.  1270. 

64.  Draw  the  first  line  at  any  angle;  and  by  the  preceding 
problem,  draw  the  two  intermediate  lines,  making,  with  the 
horizontal  line,  angles  equal  to  that  made  by  the  first  line.  The 
pupil  will  have  no  difficulty  with  the  oblique  line  at  the  other 
extremity  of  the  horizontal  line. 

65.  Proceed  as  in  the  second  part  of  64,  using  a  horizontal 
3-inch  line  for  the  diagonal,  drawing  from  one  end  a  2|-inch 
oblique  line  running  upward  at  any  angle,  and  from  the  other 
end  a  similar  line  running  downward  at  the  same  angle. 

The  lines  divide  the  3-inch  diagonal  into  5  equal  parts,  each 
part  measuring  f  inch. 

66.  The  scholars  should  gradually  learn  that  it  is  unnecessary 
to  measure  the  oblique  lines,  Km  ,     ^t^ 
and  Ln  (Fig.  34).     In  setting  off                            5. 
the  divisions  ATI,  1  2,  etc.,  La,  ah,            t 

etc.,  it  will  be  found  unnecessary  to  K^=— ^ -^^L 

use  a  definite  measure;  all  that  is  . ^^ 

required  is  that  they  be  equal.    This 

equality  is  obtained  by  opening  the      n- 

compasses  slightly  and  marking  off 

the  divisions  with  them.     There  is  no  need  of  completing  the 


Fig.  34. 


268  MANUAL   FOR  TEACHERS 

rhomboid,  nor  of  drawing  the  lines  4a,  36,  2c,  etc.  Place  the 
ruler  on  4  and  on  a,  and  use  a  short  line  on  KL  to  mark  off  one 
division;  etc. 

It  is  expected  that  the  teacher  will  not  give  this  method  at  the  outset. 

To  draw  a  line  exactly  f  of  an  inch  in  length,  a  construction 
line,  3  inches  long,  KL  (Fig.  34),  is  drawn  lightly,  also  Kw,  and 
Ln  of  indefinite  length.  Only  the  first  division  is  required  on 
Ktn.  Placing  the  ruler  on  1  and  c?,  mark  off  on  KL  f  inch, 
and  darken  this  portion 

One-seventh  of  a  5-inch  line  =  -f-  inch. 

67.  When  the  base  of  a  right-angled  triangle  measures  3 
inches,  and  the  perpendicular  measures  2  inches,  the  hypotenuse 
will  measure  V3'  -f-  2'  =  V9  +  4  =  Vl3  inches. 

68.  A  line  Vl3  inches  long  is  drawn  by  constructing  a  right- 
angled  triangle  having  a  3-inch  base  and  a  2-inch  perpendicular. 
The  hypotenuse  is  the  required  line. 

A  base  of  2  inches  and  a  perpendicular  of  1  inch  give  a  hypot- 
enuse of  V5  inches. 

A  hypotenuse  of  4  inches  and  another  side  of  3  inches  give  a 
remaining  side  of  VV  inches.  For  the  construction  of  this  tri- 
angle, see  Problem  61. 

69.  The  side  opposite  the  angle  of  30°  is  one-half  as  long  as 
the  side  opposite  the  angle  of  90°. 

71.  The  chord  of  an  arc  of  60°  is  2  inches  long  ;  the  chord  of 
an  arc  of  180°  is  the  diameter,  and  is  4  inches  long.  The  chord 
of  an  arc  of  300°  is  2  inches  long. 

72.  The  perpendicular  "erected"  at  one  end  of  the  chord 
should  be  turned  downwards  if  the  chord  is  drawn  above  the  cen- 
ter of  the  circle. 

The  perpendicular  and  the  diameter  meet  at  the  circumference. 
See  Problem  61. 

73.  The  triangle  will  retain  its  shape,  because  only  one  tri- 
angle can  be  drawn  with  sides  of  a  given  length.  Problem  59. 
The  rectangle  will  not  retain  its  shape,  because  an  indefinite  vari- 


NOTES   ON   CHAPTER   SIXTEEN  269 

ety  of   parallelograms  can  be   constructed  having  their  corre- 
sponding sides  equal,  each  to  each.     See  Exercise  66,  Art.  1266. 

77.  From  Problem  15,  the  pupils  have  learned  that  a  perpen- 
dicular bisecting  a  chord,  passes  through  the  center.  If  there 
are  two  lines  passing  through  the  center,  the  latter  must  be 
located  at  their  intersection. 

78.  To  inscribe  a  circle  in  any  triangle,  draw  lines  bisecting 
the  three  angles.  See  problem  28.  The  intersection  of  these 
three  lines  will  be  the  center  of  the  circle. 

In  practice,  only  two  hnes  are  drawn ;  but  the  third  serves  to  test  the 
accuracy  of  a  pupil's  work. 

79.  The  sides  of  an  inscribed  triangle  are  chords  of  the  cir- 
cumscribing circle.  In  Problem  77,  we  have  learned  that  the 
intersection  of  two  perpendiculars  that  bisect  chords,  locates 
the  center  of  the  circle.  With  this  center,  and  with  a  radius 
equal  to  the  distance  from  the  center  to  the  vertex  of  any  angle 
of  the  triangle,  draw  the  circle. 

80.  The  larger  the  circle  the  better  for  beginners.  The 
average  scholar  may  consider  it  necessary  to  draw  two  adjacent 
chords,  but  every  purpose  will  be  served  by  cutting  the  circum- 
ference by  three  short  lines  to  mark  the  boundaries  of  two 
adjoining  arcs.  The  bisection  of  these  arcs  by  perpendiculars 
will  locate  the  center  of  the  circle. 

82.  Use  a  cup  to  draw  the  arc.  Divide  it  at  random  into  two 
parts.     Bisect  each,  as  above.  q  JD 

83.  The  altitude  is  2  in.  ^'  '  ^^^^  4^ 
Problem  70. 

84.  Experienced  draughts- 
men save  time  by  changing  the 
radius  as  infrequently  as  possi- 
ble. To  construct  a  square  on 
a   3-inch   line  AJB  (Fig.  35),  '''''■  '^• 

take  a  radius  of  3  inches,  and  with  A   as   a   center  draw  the 
arc  ^m.     With  ^  as  a  center,  mark  off  JBn  =  60°.     Bisect  Bn, 


270  MANUAL   FOR  TEACHERS 

using  the  same  radius  to  obtain  the  arcs  intersecting  at  o,  with 
B  and  n  as  centers.  BX=  30°  ;  place  the  compasses  on  X,  and 
cut  the  first  arc  at  (7,  which  makes  XC  60°,  and  BC  90°.  C  is 
the  third  corner  of  the  square.  Using  B  and  (7  as  centers,  an  I 
with  a  radius  of  3  inches,  draw  arcs  intersecting  at  D,  the  fourth 
corner  of  the  square.  ^ m^ 

86.  We  have  found  in  Problem  26, 
that  constructing  an  equilateral  tri- 
angle on  each  side  of  an  inscribed 
equilateral  triangle  gives  a  circum- 
scribed equilateral  triangle.  Locate 
m,  n,  and  o,  120°  apart.  With  each  as 
a  center,  and  with  a  radius  equal  to 
mo,  draw  arcs  intersecting  at  A,  B, 
and  C.     Do  not  draw  the  triangle  7)ino,  shown  in  the  figure. 

87.  See  Problem  41,  second  part. 

89.  See  Arithmetic,  Art.  1089  (Fig.  2). 

90.  Construct  a  right-angled  triangle  having  a  base  and  a 
perpendicular  measuring  2  inches  and  3  inches,  respectively. 
The  hypotenuse  will  be  the  side  of  the  required  square. 

Do  not  construct  squares  on  the  other  two  sides. 

91.  The  hypotenuse  is  3  inches,  etc.     See  Problem  68. 

92.  The  square  constructed  on  the  hypotenuse  of  a  right- 
angled  triangle  whose  base  and  perpendicular  measure  3  inches 
each,  will  be  double  the  area  of  a  3-inch  square. 

When  the  3-inch  square  is  constructed,  measure  the  length  of 
the  diagonal  (without  drawing  it) ;  and  on  a  line  of  this  length 
as  a  base,  construct  the  required  square. 

93.  Four.     See  Fig.  36. 

94.  Nine  ;  sixteen  ;  twenty-five. 

95.  1,  Ij^,  2  inches. 

96.  3,  4J,  6  inches. 

97.  The  radius  of  the  second  circle  is  2  inches. 


NOTES   ON   CHAPTER   SIXTEEN 


271 


Fig.  37. 


98.  A  square  described  on  the  base  of  an  isosceles  rigbt- 
angled  triangle  will  be  one-half  the  size  of  the  square  described 
on  the  hypotenuse.  A  =  a-\-a^=2a', 
a  =  ^A.  The  area  of  an  equilateral 
triangle  whose  base  is  YZ,  is  one-half 
that  of  one  whose  base  is  XY.  The 
area  of  a  circle  whose  radius  is  YZ 
(or  XZ),  is  one-half  that  of  a  circle 
whose  radius  is  XY. 

Note.  —  If  the  square  A  in  Fig.  37  is 
drawn  above  XY  instead  of  below  it,  the 
ratio  of  the  large  square  to  that  of  each 
small  one  will  be  more  apparent. 

Construct  an  isosceles  right-angled  triangle  having  a  hypote- 
nuse of  2  inches  (Problem  52).  The  base  of  this  triangle  will  be 
the  radius  of  the  required  circle. 

The  area  of  a  circle  of  2  inches  radius  =  2^  X  tt  =  4  tt. 

The  radius  of  the  other  circle  =  V2 ;  its  area  is  (V2)'''  X  tt 
=  2  TT,  which  is  one-half  of  4  tt. 

99.  The  side  of  the  required  triangle  measures  V2  inches. 
See  98. 

100.  Cab  (Fig.  38)  is  one-sixth  of  an  inscribed  regular  hex- 
agon, and  CAB  is  one-sixth  of  a  circum-  a  X  B 
scribed  regular  hexagon.  Calling  the 
radius  of  the  circle  2  inches,  the  altitude 
CX  of  the  large  triangle  is  2  inches.  Since 
ab  is  2  inches,  ax='[  inch  -,  Ca  =  2  inches. 
In  the  right-angled  triangle  Cax,  Cx  -f  ax 
---O?;  CS'-f  1=4;  C'x'  =  4-1  =  3;  Cx 
=  Vs.  The  areas  of  the  two  triangles  will 
be  proportional  to  the  squares  of  their  respective  altitudes. 
C^'  =  3  and  GX^  =  4.  That  is,  the  area  of  the  larger  triangle  is 
1-^^  times  the  area  of  the  smaller ;  hence  the  area  of  the  circum- 
scribed hexagon  is  \\  times  the  area  of  the  inscribed  hexagon. 

The  area  of  the  circumscribed  square  is  double  the  area  of  the 


272 


MANUAL   FOR  TEACHERS 


inscribed  square;  and  the  area  of  the  circumscribed  equilateral 
triangle  is  four  times  the  area  of  the  inscribed  equilateral 
triangle. 

1271.  3.  At  the  middle  point  Xof  the  perpendicular  of  the 
right-angled  triangle  (Fig.  39),  cut  Xtyi  parallel  to  the  base. 
Re-arranging  the  two  parts  gives  a  rectangle  whose  dimensions 
are  4  inches  and  \\  inches.     Figs.  40  and  41  show  how  the 

A  A  A 


Fig.  39.  Vva.  40.  Fig.  41. 

other  two  triangles  are  divided  to  make  rectangles  of  the  same 
dimensions. 

4-6.    See  56-59,  Art.  1270. 

7.  The  two  triangles  have  two  sides  of  one,  AC  and  BC, 
equal  to  two  sides  of  the  other,  CE  and  CD\  and  the  angle 
ACB  equal  to  its  opposite  angle  DCE\  hence  the  third  side, 
DE,  of  one  triangle  is  equal  to  the  third  side,  AB,  of  the  other. 

9.  The  area  of  the  larger  triangle  is  four  times  that  of  the 
smaller. 

11.  Another  method  of  dividing  a  line  into  equal  parts.  See 
Problem  66,  Art.  1270. 

1273.  Much  interest  is  added  to  this  work  by  employing  the 
method  here  given,  in  calculating  heights  and  distances  in  the 
neighborhood  of  the  school.  A  comparison  between  the  results 
obtained  by  calculations  and  those  obtained  by  actual  measure- 
ments, will  be  useful  in  teaching  the  pupils  the  necessity  of  great 
accuracy  in  their  preliminary  work. 

2.  The  hypotenuse  of  each  triangle  represents  a  ray  of  light 
from  the  sun.  etc. 


NOTES   ON   CHAPTER   SIXTEEN 


273 


ft. +  4^ 


1- ft.  =  162  ft.  Ans. 
RN'.:  FQ.MJST;  6:  (120 +  6) 


hf  =157i  ft. 


4:ifiV.    MJY-- 


3.  Fr=110ft. 

4.  CD.DE'.:  CB.BA;  50  :  90  : :  1200  :  ^^  ;  ^^  =  2160ft. 
The  width  of  the  river  =  2160  ft.  -  100  ft.  =  2060  ft.  Ans. 

5.  CD=CG-DG  =  6  ft.-4  ft.  =  2  ft.;  EII=lll  ft. 
+  3  ft.  =  180  ft ;  AH=  120  ft. ;  AB  =  120  ft.  +  4  ft.  =  124  ft. 

6.  The  two  triangles  are  similar,  because  the  angles  of  one  are 
equal  to  the  angles  of  the  other.  Angle  D  =  angle  A  =  90°. 
The  two  angles  at  C  are  vertical  angles,  and,  therefore,  equal  to 
each  other.  The  remaining  angle  J5  must  be  equal  to  the 
remaining  angle  U,  because  the  sum  of  the  angles  in  each 
triangle  is  the  same. 

CD.DE::  AC:AB; 
3.25:     5    ::  (12 -3.25):  ^5. 

7.  lOi- :  (lOi  +  195  +  15)  :  :  (12  -  41)  :  A/. 
^=15 

8.  BP 
84  ft.  Ans, 

9.  The  tree  is  3  ft.  X  36,  or  108  ft.  high.  Ans. 
because  angle  C=  angle  B  =  45°. 

10.  AC=  AB.  A  line  CB  makes  an  angle  of  45°  with  AC, 
angle  ^  =  90° ;  angle  ABC=  45°. 

Note.  —  In  the  succeeding  problems,  the  triangle  and  the  protractor  may 
be  employed  when  necessary. 

1274.  1.  The  circumference  of  the  circle,  or  360°  =  4  in. 
X  3.1416.  The  arc  of  60°  is  J  of  the  circumference ;  the  arc 
of  120°  =  i  of  it ;  etc. 

2.    Draw  the  circle  and  the  various  chords. 
=  chord  of  300°  =  radius  =  2  inches ;  the  chord 
of  180°  =  diameter  =  4  inches. 

If  the  pupils  draw  the  chord  of  120°,  ZX 
(Fig.  42),  they  will  find  that  it  bisects  the  ra- 
dius YC,  making  MC  1  inch.  CX-=  2  inches  ; 
therefore  MX  =  ^¥^^  =  V3  =  1.732  +  ; 
ZX=  1.732  +  in.  X  2  =  3.464  +  in.  =  chord  of 
120°  =  chord  of  240°. 


AC=AB, 


The  chord  of  60° 


274  MANUAL   FOR  TEACHERS 

Diagrams  should  be  employed,  unless  the  pupils  can  do  satis- 
factory work  without  them.  Since  the  arc  of  180°  is  three  times 
as  long  as  the  arc  of  60°,  the  more  careless  members  of  the  class 
may  jump  to  the  conclusion  that  the  chord  of  180°  is  three  times 
as  long  as  the  chord  of  60°.  This  mistake  cannot  be  made  if  the 
circle  is  constructed,  and  the  chords  are  drawn  and  measured. 

5.  See  Exercise  98,  Art.  1269. 

6.  Each  side  of  the  hexagon  measures  1  inch ;  the  perimeter 
=  1  inch  X  6  =  6  inches.  Ans. 

Circumference  =  2  in.  X  3.1416  =  6.2832  inches.  Ans. 

7.  The  pupil  should  use  the  ruler  to  ascertain  the  length  of 
the  apothem,  which  is  about  -J-  in.  It  can  be  calculated  as 
follows :  — 

Cb  (Fig.  38)  is  the  radius  =  1  inch ;  bx  =  one-half  the  side 
of  the  inscribed  hexagon  =  ^  inch ;  Cx  =  apothem.  Cx  =  Cb  — 
bx=l-\  =  .1b;  Cx  =  -yJTfb  =  .866+,  or  nearly  J. 

8.  The  base  of  each  triangle  measures  one  inch,  so  that  AB 
(Fig.  43)  =  3  in.     ^X  (apothem)    x 

=  J  in.  nearly. 

9.  The  base  of  each  triangle 
measures  about  f  in.  so  that  the  A 
base   of  the   rectangle    (the   half  ^^^'  ^' 
perimeter)  will  be  nearly  3  inches  and  the  apothem  about  \^  in. 
Area  about  3  X  ||-  sq.  in.  =  about  2^  sq.  in. 

10.  AB  X  AX=  3  X  J  =  2f .  Ans.  2f  sq.  in. 

11.  See  9. 

12.  The  perimeter  of  a  16-sided  polygon  will  be  greater  than 
that  of  an  octagon.  The  16-sided  polygon  will  have  the  greater 
apothem. 

13.  As  the  number  of  sides  increases,  the  perimeter  approaches 
more  and  more  closely  the  circumference,  6.2832  inches;  and  the 
apothem  approaches  the  radius,  1  inch. 


NOTES   ON   CHAPTEE   SIXTEEN  275 

14.  The  base  of  the  rectangle  will  be  3.1416  inches,  one-half  the 
perimeter  (circumference)  ;  the  apothem  will  be  1  inch  (radius). 
Area  =  3.1416  sq.  in.  Ans. 

15.  One-half  the  circumference  =  2  X  3.1416  =  6.2832,  multi- 
plied by  the  radius,  2,  gives  answer  in  square  inches,  12.5664  sq.  in. 

16.  78.54  sq.  in.  Ans. 

17.  3.1416  sq.  in.  Ans. 

18.  314.16  sq.  in. -^6  =  Ans. 

19.  Subtract  from  the  area  of  the  outer  circle,  113.0976  sq. 
in.,  the  area  of  the  inner  circle,  28.2744  sq.  in.  Ans.  84.8232* 
sq.  in. 

1282.  Bight  prisms,  cylinders,  etc.,  are  meant  when  the  word 
oblique  is  not  used. 

1.  See  Arithmetic,  Art.  818,  Problem  20.  The  upper 
squares  need  not  be  drawn,  as  only  the  convex  surface  is  required. 

2.  Three  rectangles  and  two  triangles.     See  1. 

3.  Three  rectangles,  each  3  inches  high,  bases  1,  1^,  and  2 
inches,  respectively. 

5.  A  hollow  paper  cylinder,  without  bases,  can  be  opened 
out  into  a  rectangle  whose  base  is  the  circumference  of  the  base 
of  the  cylinder,  and  whose  height  is  the  altitude  of  the  cylinder. 

6.  The  slant  height  of  a  square  pyramid  is  the  distance  from 
the  apex  to  the  center  of  one  side  of  the  base.  This  problem 
requires  the  pupil  to  draw,  side  by  side,  four  isosceles  triangles, 
the  base  of  each  being  2  inches  and  the  altitude  3  inches.  After 
constructing  the  first  (Fig.  44),  by  erect- 
ing a  3-inch  perpendicular  at  the  center 
of  a  2-inch  base,  he  should  use  B"  and 
A  as  centers,  and  radii  equal  to  ^"^'and 
AB"  to  locate  B"\  On  AB'  construct  a  B'"^ 
third  triangle,  using  A  and  B'  as  centers 
and  the  radii  previously  given.  Upon  one  ^t  '^Ji^ 
side  of  this  triangle  construct  a  fourth.  Fig.  44. 


276  MANUAL   FOR  TEACHERS 

The  pupils  can  discover  for  themselves  the  method  of  drawing 
geometrically  the  required  convex  surface. 

The  altitude  should  be  carefully  measured.  It  will  be  equal 
in  length  to  the  perj^pndicular  of  a  right-angled  triangle  whose 
hypotenuse  is  the  slant  height  of  a  pyramid,  3  inches ;  and  whose 
base  is  the  distance  from  the  center  of  one  edge  of  the  base  of 
the  pyramid  (the  foot  of  the  slant  height)  to  the  foot  of  the  alti- 
tude, 1  inch.  Ans.  V8  inches  =  2.83  inches  nearly  =  about  2|J 
inches. 

7.  The  area  of  each  convex  face  of  a  regular  pyramid  is 
found  by  multiplying  one  side  of  the  base  by  one-half  the  slant 
height;  therefore,  the  area  of  all  the  faces  forming  the  convex  sur- 
face, is  obtained  by  multiplying  the  sum  of  all  the  sides  of  the  base, 
that  is,  the  perimeter  of  the  base,  by  one-half  the  slant  height. 

8.  The  pupil  requires  the  slant  height  in  order  to  proceed  as 
in  6  ;  and  he  should  obtain  it  by  drawing  it  rather  than  by  cal- 
culating it.  He  should  be  able  to  see  that  the  altitude  is  a  line 
drawn  from  the  vertex  to  the  center  of  the  base,  and  that  its  foot 
is  1  inch  distant  from  the  foot  of  the  altitude.  Constructing  a 
right-angled  triangle  whose  base  is  1  inch,  and  whose  perpen- 
dicular is  3  inches,  will  give  a  hypotenuse  equal  to  the  required 
slant  height. 

Some  scholars  will  bring  in  a  prism  whose  slant  height  is  3 
inches.  The  mistake  should  be  pointed  out,  but  not  the  mode 
of  correcting  it;  and  a  pyramid  of  the  required  dimensions 
should  be  insisted  upon. 

1283.  When  the  diameter  of  the  base  of  a  cone  is  2  inches, 
the  arc  BDC,  which  forms  the  circumference  when  folded, 
measures  27r  inches,  or  6.2832  inches. 

Note.  —  The  Greek  letter  ir  (pi)  represents  3.1416. 

9.  The  semi-circumference  of  paper  =  3ir  inches,  which  is  the 
circumference  of  base  of  cone.  The  diameter  of  base  of  cone 
—  3  TT  inches -^  TT  =  3  inches.      The  radius  of  base  =  1}  inches. 


r 


NOTES   ON   CHAPTER   SIXTEEN 


277 


The  slant  height  =  radius  of  paper  =  3  inches  =  diameter  of  base 
of  cone  =  twice  radius  of  base  of  cone. 

10.  The  area  of  any  sector  of  a  circle  =  radius  X  i  length  of 
arc.  The  arc  when  folded  becomes  the  circumference  of  base, 
and  the  radius  becomes  the  slant  height ;  so  that  convex  surface 
=  i  circumference  X  slant  height  =  circumference  X  i  slant 
height. 

11.  Length  of  arc  of  90°  =  -^  circumference  = -^  of  Gtt  inches 
=  1^  TT  inches.  This  is  the  circumference  of  the  base  of  the  cone, 
which  makes  the  diameter  =  1|-  tt  inches  -^  tt  =  1|-  inches.  The 
slant  height  is  3  inches. 

Length  of  arc  of  60°  =  ^  circumference  =  |-  of  6  tt  inches 
=  TT  inches.  The  diameter  of  the  base  of  the  cone  =  tt  inches  ^  tt 
=  1  inch;  slant  height,  3  inches. 

12.  The  circumference  of  the  base  of  the  cone  =  3ir.  This 
equals  the  length  of  the  arc  of  the  required  sector.  If  the  slant 
height  is  6  inches,  the  circumference  of  which  the  sector  is  a  part 
=  10  TT.  The  arc  of  the  sector  is,  therefore,  y^^  of  the  circumfer- 
ence ;  and  its  length  is  -^-^  of 
360°  =  108°. 

13.  XY  represents  the 
base  of  the  pyramid;  and 
AC,  its  altitude.  The  slant 
height  of  two  faces,  AM 
(Fig.  45),  is  the  hypotenuse 
of  a  right-angled  triangle,  base 
IJ  in.,  perpendicular  4  in. 
Using  this  as  the  perpendicu- 
lar of  a  new  triangle,  with  a 
base  MY,  gives  as  the  hy- 
potenuse AY  (Figs.  45  and 
46),  one  of  the  edges. 

To  draw  the  development, 
take  AY  as  a  radius,  and  draw  an  arc 


Fig.  45. 


On  this  lay  off  succes- 


278 


MANUAL   FOR  TEACHERS 


sive  chords  of  3  in.,  2  in.,   3   in.,  and   2  in.,  connecting  the 
extremity   of  each   with   the   center  A,     These   four  triangles 


constitute  the  convex  faces  of  the  pyramid.     On  one  of  them, 
construct  a  rectangle  3  inches  by  2  inches  for  the  base. 

14.  In  this  pyramid,  AC  (Fig.  45)  measures  12  inches,  C3I 
measures  -^  of  18  inches,  or  9  inches,  making  AM,  one  slant 
height,  =  Vl44  -f  81  inches  =  15  inches.  The  other  slant  height, 
drawn  to  the  center  of  OF,  =  Vl44  -f-  25  inches  =  13  inches. 


The  pupils  should  be  encouraged  to  construct  stout  paper 
pyramids  and  cones  of  required  dimensions,  making  the  necessary 
calculations  themselves.  The  previous  fourteen  problems  will 
present  no  difficulty  whatever  to  pupils  that  are  interested.  As 
examples  in  dry  calculation,  they  may  prove  somewhat  tiresome. 
Many  scholars  will,  of  themselves,  construct  models  of  solids  much 
more  complicated  than  the  foregoing. 

1284.  If  there  are  no  solids  at  hand,  the  pupils  should  con- 
struct a  supply  of  paper  ones,  or  make  them  of  modeling  clay, 
turnips,  etc.  Drawings  are  not  sufficient  for  effective  instruc- 
tioD. 


NOTES   ON    CHAPTER   SIXTEEN  279 

15.  The  short  method  of  ascertaining  the  convex  surface  should 
not  be  given  until  18.  Each  of  the  convex  faces  is  a  trapezoid, 
whose  parallel  sides  measure  4  inches  and  8  inches,  respectively, 
the  altitude  being  10  inches. 

16.  Two  inches  apart,  draw  two  parallel  lines,  AB  and  CD 
(Fig.   47),   measuring   1   inch    and   2  inches, 

respectively,    a    2-inch    perpendicular,    ^Y,  i 

connecting  the  middle  point  of  each.     Draw  /;\ 

AC  and  BD,  and  produce  them  until  they  j  \  \ 

meet  in  M.     With  this  as  a  center,  and  MO 
as  a  radius,  draw  an  arc ;    on  which   three  / 

other  chords  equal  to  CD  are  laid  off,  as  in  ^p 

Fig.  46,     With  jif  as  a  center  and  a  radius  / 

MA,  lay  off  another  arc,  on   which    chords  / 

are  laid  off  equal  to  ^^;  etc.     See  Arithme-         / 
tic.  Art.  1296.  /  jir, 

^ X 

17.  The  entire  surface  =  convex  surface  +  ■ 

4  sq.  ft.  +  9  sq.  ft.  ^'"-  '"■ 

18.  The  pupils  can  readily  understand  this  rule,  using  the 
frustums  of  problems  15  and  17  as  illustrations. 

19.  The  pupil  should  first  locate  the  apex  of  the  cone.  This 
he  can  do  by  following  the  method  shown  in  16  ;  making  AB 
(Fig.  47)  2\  inches;  CD,  1^  inches;  and  AC,  2  inches.  This 
makes  MA  5  inches,  the  slant  height  of  the  whole  cone.  The 
circumference  of  the  base  of  the  cone  —  2^7r.     The  slant  height, 

5  inches,  is  the  radius  of  the  required  sector,  its  circumference 
being  10  tt.  2|-7r,  the  length  of  the  arc  of  the  sector,  being  one- 
fourth  of  10  TT,  shows  that  the  required  sector  is  a  quadrant. 

20.  The  number  of  square  inches  in  the  convex  surface  =  [cir- 
cumference (perimeter)  of  upper  base  (9  X  3.1416)  +  circumfer- 
ence of  lower  base  (6  X  3.1416)]  X  ^  slant  height  (2).  Adding  to 
this,  the  area  of  the  bottom  (3')  9  sq.  in.  X  3.1416  (Art.  1124, 
2)  gives  the  number  of  square  inches  of  material  required. 


280  MANUAL    FOR   TEACHERS 

22.    Circumference  of  upper  base  =    6x3. 1416 

Circumference  of  lower  base  =  10  X  3.1416 


One-half  sum  =   8  X  3.141G 


Multiplying  by  slant  height  gives  48  X  3.1416 

Add  to  this  the  area  of  the  upper  base,  9  X  3.1416 

And  the  area  of  the  lower  base,  25  X  3.1416 

Total  in  square  yards,  82  X  3.1416 

or  ( [(3  +  5)  X  6]  -f  9  +  25)  x  3.1416. 

23.  FA:   UC   ::Q:S 

x:x  +  9  ::  6:8 
8a:  =  6a;  +  54 
2a:=:54 
a; -27.  Ans.  27  ft. 

The  slant  height  of  the  whole  cone  =-27  ft.  +9  ft.  =36  ft. 

24.  The  convex  surface  of  the  whole  cone  =  ^  (8x3.1416 
X  36)  sq.  ft. ;  of  the  part  cut  off =i  (6  X  3.1416  X  27)  sq.  ft. 

1287.  A  sphere  (a  croquet  ball,  for  instance)  and  a  hemisphere 
should  be  used  to  illustrate  these  problems.  On  the  plane  face 
of  the  latter  can  be  drawn  the  lines  AD,  FG,  HI,  CI,  etc. ; 
while  on  the  curved  face  can  be  drawn  HYI,  FXQ^  etc. 

25.  ^  of  25,000  miles. 

26.  IH^  chord  of  60°  of  the  great  circle  =  radius  of  the 
<^ieat  circle  =  4000  miles.     Z5  =  J  of  111=  2000  miles. 

27.  The  diameter,  HI,  of  the  small  circle  is  \  diameter  FG 
of  the  great  circle ;  the  circumference  HYI=  \  of  25,000  miles, 
or  12,500  miles. 

28.  The  length  of  a  degree  of  longitude  on  the  60th  parallel 
is  about  one-half  of  the  length  of  a  degree  on  the  equator. 
(See  Art.  995,  Problem  10.) 


NOTES   ON   CHAPTER   SIXTEEN 


281 


29.  On  the  plane  face  of  the  hemisphere  suggested  above 
(Art.  1287),  draw  diameters  AD  and 
FQ  at  right  angles ;  and  45°  from  G, 
a  chord  NM  parallel  to  FG.  (This 
chord  will  not  bisect  AC.)  As  MOG 
-  45°,  WCM-=  45° ;  and  the  triangle 
WCM  is  a  right-angled  isosceles  tri- 
angle, and  WM'^^^CM'')  WM= 
.lOnCM.  (See  Art.  995,  Problem 
12.)  If  CG  measures  4000  miles, 
CM=  4000  miles,  and 
WM-=  V8,000,000  =  2828.4  miles. 

1289.  The  pupils  have  already  learned  that  the  volume  of  a 
rectangular  prism  is  equal  to  the  area  of  the  base  multiplied  by- 
its  altitude ;  these  problems  are  intended  to  show  that  the  same 
is  true  of  all  prisms  and  of  the  cylinder  (6). 

1292.  8.  The  volume  of  the  frustum  is  obtained  by  deduct- 
ing the  volume  of  the  part  cut  off  from  the  volume  of  the  whole 
pyramid  (Problem  7).     The  rule  is  given  later. 

10.  Fig.  49  gives  the  method  of  calculating  the  slant  height. 
The  illustration  shows  a  sec- 
tion of  the  frustum  formed 
by  passing  a  plane  through 
the  center  of  the  frustum 
perpendicular  to  the  base. 
The  center  of  the  upper  base 
of  the  frustum  of  a  right  pyr- 
amid is  directly  above  the 
center  of  the  lower  base,  so 
that  a  perpendicular  let  fall 
from  A  will  fall  on  CD  at  a 
point  Jl,  5  in.  from  C.  In 
the  right-angled  triangle  AXC,  AX- 
=  ^"X'+^'  =  1444-25  =  169;  AC- 


5  in, 


Fig.  49. 

altitude  =  12  in. 
=  Vl69  =  13. 


AC^ 


XX 


NOTES  ON   THE   APPENDIX 

1306.  A  person  that  contracts  to  receive  a  rate  of  interest 
greater  than  is  permitted  by  law,  is  liable  to  a  penalty,  except  in 
Connecticut.  In  Delaware,  Minnesota,  etc.,  the  penalty  is  the 
forfeiture  of  the  contract ;  in  New  York,  the  forfeiture  of  the 
contract,  $1000  fine,  and  6  months'  imprisonment;  in  Indiana, 
Kansas,  Kentucky,  Maryland,  Michigan,  Mississippi,  Ohio,  Penn- 
sylvania, Tennessee,  Vermont,  Virginia,  and  West  Virginia,  the 
forfeiture  of  the  excess  of  interest ;  in  North  Carolina,  the  forfeiture 
of  double  the  amount  of  interest ;  in  Georgia  and  New  Hamp- 
shire, the  forfeiture  of  three  times  the  excess  of  interest ;  in  Ala- 
bama, Delaware,  Florida,  Illinois,  Iowa,  Louisiana,  Missouri, 
Nebraska,  New  Jersey,  South  Carolina,  Texas,  and  Wisconsin, 
the  forfeiture  of  all  interest ;  in  Arkansas  and  Oregon,  the  for- 
feiture of  principal  and  interest. 


1307.     3.    Amount  of  $  1000,  June  1,  1896.  to  June  1,  1897 
Amount  of  $150,  Sept.  16,  1896,  to  June  1,  1897,  8^  mo. 

Due  June  1,  1897 

Interest  to  settlement,  Apr.  16,  1898,  lOJ  mo.   . 

Amount  Apr.  16,  1898         .... 
Amount  of  $50,  Sept.  16,  1897,  to  Apr.  16,  1898,  7  mo. 

Balance  due 


11,060.00 
166.37 

$903.63 
47.44 

$951.07 
51.75 

$899.32 


4.    Amount  of  $500,  July  25,  1896,  to  Apr.  1, 

1897,  8  mo.  6  da. ^  $520.50 

Amount  of  $100,  Sept.  18,  1896,  to  Apr.  1,  1897, 

6  mo.  13  da $103.22 

Amount  of  $200,  Feb.  5,  1897.  to  Apr.  1,  1897,  1  mo. 

26  da 201.86  305.08 


Balance  due 


282 


$215.42 


NOTES   ON   THE   APPENDIX 


283 


$985.99 

235. 
$750.99 
35.54 
$786.53 

255.33 


5.    Amount  of  $870.50,  Jan.  2, 1894,  to  March  18,  1896,  2  yr, 

2  mo.  16  da 

Less  payments  of  $  35  and  $  200 

New  principal  ....... 

Interest  March  18,  1896,  to  Jan.  2,  1897,  9  mo.  14  da.  . 

Amount 

Amount  of  $250,  Aug.  24,  1896,  to  Jan.  2,  1897,  4  mo.  8  da, 

Due  at  maturity $531.20 

Below  will  be  found  answers  to  the  partial  payments  examples 
of  Chapters  XIII  and  XIV,  according  to  tlie  Connecticut  rule. 

Note.  —  Although  the  legal  rate  in  Connecticut  is  6%,  the  rates  given  in 
the  examples  should  be  used. 

Art.  1008.  $70.51.  Art.  1009.  1.  $224.22.  2.  $261.21. 
3.  $1278.15.  Art.  1011.  4.  $771.24.  5.  $899.32  (see  Art. 
1307,  3).  Art.  1013.  5.  $1088.31.  Art.  1015.  7.  $700.50. 
Art.  1023.  7.  $1232.26.  Art.  1051.  7.  $77.07.  Art.  1090. 
3.  $649.13.     4.   $224.64.     Art.  1107.    2.  $1089.64. 

The  time  in  each  of  the  preceding  examples  was  found  by  com- 
pound subtraction,  taking  30  days  to  each  month.  In  the 
examples  in  Art.  1110,  the  Connecticut  rule  has  been  followed. 
In  these,  the  time  is  taken  in  days.     (See  Art.  1111.) 


1308.     7.    Principal $700.00 

Interest  to  June  15,  1897,  2  yr.         .         .         ,         .         .  84.00 

1  year's  interest  on  $42,  unpaid  interest           .         .         .  2.52 

Amount  June  15,  1897 $786.52 

Amount  of  $20,  Nov.  15,  1895,  to  June  15,  1897     .         .  $21.90 

Amount  of  $80,  Feb.  15,  1897,  to  June  15,  1897     .        .  81.60       103.50 

New  principal  June  15,  1897     ....  $683.02 

Interest  to  Oct.  15,  1899,  2  yr.  4  mo 95.62 

Interest  for  1  yr.  4  mo.  on  $  40.98,  unpaid  interest  .         .  3.28 

Interest  for  4  mo.  on  $40.98,  unpaid  interest  ...  .82 

Amount  Oct.  15,  1899 $782.74 

Amount  of  $  15,  Sept.  15,  1898,  to  Oct.  15,  1899      .        .  15.98 

Due  Oct.  15,  1899 $766.76 

Note.  —  If  four  places  of  decimals  are  used,  the  answer  to  6  is  $367.6442, 
or  $367.64  +  ;  to  7,  $  766.7658,  or  $  766.77  - . 


284 


MANUAL   FOR  TEACHERS 


The  teacher  that  wishes  other  examples  of  this  kind  can  use  9  and  10  of 
Art.  1309,  and  12, 14,  and  15  of  Art.  1310,  the  answers  to  which  are  as  fol- 
lows: Art.  1309.  9.  $1734.  10.  $738.29.  Art.  1310.  12.  $1901.18.  14. 
$3432.20.     15.  $1010.95. 

1309.  Pupils  in  New  Hampshire  should  not  be  taught  the  pre- 
ceding method,  but  should  be  confined  to  the  rule  laid  down  by 
the  courts  of  their  own  state. 

An  examination  of  8  will  show  the  manner  of  ascertaining  the  balance. 
The  interest  for  1  year  is  $36.  As  no  interest  is  due  except  that  which 
accrued  during  the  year,  and  as  the  sura  paid  is  less  than  the  interest  due, 
no  interest  is  allowed  on  the  payment  of  $30,  made  during  the  year.  This 
payment  being  $6  less  than  the  interest  due  at  the  end  of  the  year,  the  in- 
terest on  $6  is  added  to  the  interest  on  the  principal  at  the  end  of  the  next 
year,  making  a  total  of  $642.36  then  due  less  the  amount  of  $100  for  11 
months.  Interest  is  allowed  on  the  $  100  payment  because  it  is  in  excess  of 
the  interest  then  due. 


9.  Principal 

Annual  interest  due  May  1,  1897 
Payment  (no  interest)  Oct.  1,  1896 

Balance  of  interest 
Annual  interest  due  May  1,  1898 
Interest  on  balance  of  interest,  1  yr.   . 

Amount  May  1,  1898     . 
Amount  of  $1000,  June  1,  1897,  to  May  1, 

New  principal  May  1,  1898   . 
Interest  on  $  1648  to  May  1,  1899 
Payment  (no  interest)  Nov.  1,  1898 

Balance  of  interest 
Interest  on  $  1648  May  1  to  Oct.  1 
Interest  on  balance  of  interest,  5  mo.  . 

Due  Oct.  1,  1899    . 

10.  Principal 

Annual  interest  due  Jan.  3,  1896 
Payment  (no  interest)  June  1,  1895 

Balance  of  interest 


1898 


$2500.00 


$  150.00 

100.00 

$   50.00 

150.00 

3.00 

203.00 

1 2703.00 

1055.00 

$1648.00 

$98.88 

50.00 

$48.88 

41.20 

1.22 

91.30 

$  1739.30 
$1000.00 


$60.00 
10.00 

$50.00 


NOTES   ON   THE   APPENDIX 


Amounts  brought  forward 
Annual  interest  due  Jan.  3,  1897 
Interest  on  balance  of  interest,  1  yr.    . 

Interest  due  Jan.  3,  1897 
Amount  of  payment  March  14,  1896,  9  mo.  19  da. 

Balance  of  interest 
Annual  interest  due  Jan.  3,  1898 
Interest  on  balance  of  interest,  1  yr.    . 
Annual  interest  due  Jan.  3,  1899 
Interest  for  1  year  on  $  102.52  and  1 60 

Amount  Jan.  3,  1899      . 
Amount  of  |500,  Sept.  30,  1898,  to  Jan.  3,  1899 

New  principal  Jan.  3,  1899     . 
interest  on  $  730.67  to  March  11,  2  mo.  8  da. 

Due  March  11.  1899 


1 50.00 

60.00 

3.00 

$113.00 
10.48 

$  102.52 
60.00 

6.15 
60.00 

9.75 


285 
1000.00 


238.4: 


P  1238.42 
507.75 

1 730.67 
8.28 

$738.95 


Note.  —  The  amount  of  the  payment  of  March  14  canceled  the  $3  in- 
terest on  interest  and  $7.48  additional,  leaving  $  102.52  interest  still  unpaid 
Jan.  3,  1897  on  which  two  years'  interest  is  taken  to  Jan.  3,  1899.  Two 
years'  interest  on  the  principal  is  also  taken,  and  one  year's  interest  on  the 
annual  interest  due  Jan.  3,  1898  and  unpaid.  It  will  be  noticed  that  no 
interest  is  taken  on  the  interest  upon  interest,  $3  and  $6.15.  See  14  of 
Art.  1310,  as  calculated  by  the  N.H.  rule :  — 


Principal       .... 
Interest  to  March  17, 1900,  4  yr. 
Interest  on  $  180  (3  +  2  +  1)  yr. 
Amount  of  $  20,  10  mo. 

Unpaid  interest  on  interest  . 

Interest  to  March  17,  1903,  3  yr. 

Interest  on  $  180  (3  +  3  +  3  +  3  +  2  +  1)  yi 

Total  interest  due  March  17,  1903 
Amount  of  $  1000,  6  mo.       . 

Due  March  17.  1903 


$3000.00 


$64.80 
21.00 

$43.80 
162.00 


$  720.00 


540.00 

205.80 


$  1465.80 
1030.00 


435.80 


$  3435.80 


Note.  —  Interest  is  not  taken  on  $43.80. 

As  additional  examples,  the  teacher  rnay  use  7  of  Art.  1308,  and  12,  14, 
and  15  of  Art.  1310.  The  answers  by  the  N.H.  method  are  7,  $  767.60 ; 
12,  $1901.27;  14,  $3435.80;  15,  $  1011.40. 


286 


MANUAL   FOK  TEACHERS 


1310.    Vermont  pupils  should  omit  Arts.  1307,  1308,  and  1309. 


12.  Trincipal  . 
Annual  interest  due  June  1,  1897 

Amount  June  1,  1897 
Amount  of  $  100  to  June  1,  1897 
Payment  June  1,  1897 

New  principal 
Annual  interest  to  June  1,  1899 
Amount  of  $  50,  7  mo. 

Unpaid  interest 
Interest  on  $  1705  to  settlement 
Interest  on  $  158.99  to  settlement 

Due  Oct.  1,  1899     . 

13.  Principal  Jan.  3,  1895 
Interest  due  Jan.  3,  1896 
Amount  of  $  10,  7  mo.  2  da. 

Balance  of  interest 
Interest  due  Jan.  3,  1897 

1  year's  interest  on  ^49.65  . 

Total  interest 
Amount  of  $  10,  9  mo.  19  da. 

Balance  of  interest 
Interest  on  $  1000  for  2  years 

2  years'  interest  on  $  102.15 
1  year's  interest  on  $  60 

Amount  Jan.  3,  1899 
Amount  of  $500,  Sept.  30,  1898,  to  Jan  3, 

New  principal  Jan.  3,  1899 
Interest  to  March  11,  1899  . 

Due  March  11,  1899 

14.  Principal  March  17,  1896 
Interest  to  March  17,  1900,  4  yr. 
Interest  on  yearly  interest  (3  +  2  +  1)  yr. 
Amount  of  $  20,  10  mo. 

Unpaid  interest  on  yearly  interest 


1899 


164.80 
21.00 

143.80 


$  2500.00 
309.00 


$2809.00 

$104.00 

1000.00 

1104.00 

$1705.00 

$210.74 

51.75 

$158.99 

34.10 

3.18 

196.27 

$1901.27 

$1000.00 

$60.00 

10.35 

$49.65 

60.00 

2.98 

$112.63 

10.48 

$102.15 

120.00 

12.26 

3.60 

238.01 

$  1238.01 

507.75 

1^30.26 

8.28 

1 738.54 
$3000.00 


$720.00 


NOTES   ON   THE   ArPENDIX 


287 


43.80 


162.00 


Amounts  brought  forward 
3  years'  interest  to  March  17,  1903 
Interest  on  yearly  interest,  3  yr.  on  $720 
(2  +  1)  yr.  on  $  180 

Total  interest  due  March  17,  1903 
Amount  of  $  1000  to  March  17,  1903  . 

Due  March  17,  1903 

15.    Principal  Feb.  25,  1893     . 
Annual  interest  to  Feb.  25,  1897,  4  yr. 

Amount  Feb.  25,  1897    . 
Amount  of  |400,  11  mo.  29  da.    . 

New  principal  Feb.  25,  1897  . 
Annual  interest  on  $  1089.99,  2  yr.  . 
Amount  of  $  10,  8  mo.  14  da. 

Unpaid  interest  Feb.  25,  1899 
Annual  interest  on  $  1089.99,  2  yr. 
Interest  for  2  yr.  on  $  124.30 

Amount  Feb.  25,  1901    . 
Amount  of  |400,  4  mo.  29  da.      . 

New  principal  Feb.  25,  1901  . 
Annual  interest  on  $953.99,  1  yr.  1  da. 


As  additional  examples,  7  and  9  may  be  used.     The 
Vermont  rule  is  |  766.75  ;  to  9,  $  1733.73. 


.  720.00     $  3000.00 
540.00 


205.80 


1465.80 
1030.00 


435,80 


134.72 
10.42 

124.30 

134.72 

14.91 


$3435.80 

$  1200.00 
313.92 

$1513.92 
423.93 

$  1089.99 


273.93 


$  1363.92 
409.93 

$  953.99 
57.41 

$1011.40 

answer  to  7  by  the 


1311.     1.    [($8500  +  $600)  X. 0155] +  $2 -$141.05.  Ans. 
2.    The  amount  to  be  raised  on    property  =  $2500  —  ($2  X 


150) 
mills 


275000 -$.008,  or  8 


$97.     Tax  =  $2.45 


$2200.     Rate  on  $1  =  $2200 
8  mills  on  $1,  or  1%.  A72S. 

4.  Mr.  Simmons'  grand  list  =  $  95  +  $  2 
X  97  =  $237.65.  Am. 

5.  1500 -f  2a:  =  grand  list.  The  grand  list  multiplied  by  the 
rate  gives  the  amount  to  be  raised;  (1500  + 2a;)  X  2  =  3600 ; 
3000  -\-ix  =  3600 ;  4 a;  =  600 ;  a:  =  150.     150  taxable  polls.  Ans. 


288  MANUAL   FOR   TEACHERS 

6.    Mr.  Hallock's  grand  list  =  $  120  +  $6  =  $126.     Since  his 
taxes  are  $252,  the  rate  =  $252  ^  126  =  $2. 
Let  X  =  appraised  value  of  property. 

_£_  + 400  =  grand  list. 
2  (-^  +  400)  =  total  levy  =  6800.      - 

^  +  800  =  6800. 

2a; +  80000  =  680000. 
X  =  300000. 
Appraised  value  of  property  is  $300000.  Am. 


SUPPLEMENT 


DEFINITIONS,   PRINCIPLES,   AND  RULES 

A  Unit  is  a  single  thing. 

A  Number  is  a  unit  or  a  collection  of  units. 

The  Unit  of  a  Number  is  one  of  that  number. 

Like  Numbers  are  those  that  express  units  of  the  same  kind. 

Unlike  Numbers  are  those  that  express  units  of  different  kinds, 

A  Ooncrete  Number  is  one  in  which  the  unit  is  named. 

An  Abstract  Number  is  one  in  which  the  unit  is  not  named. 

Notation  is  expressing  numbers  by  characters. 

Arabic  Notation  is  expressing  numbers  by  figures, 

Eoman  Notation  is  expressing  numbers  by  letters. 

Numeration  is  reading  numbers  expressed  by  characters. 

The  Place  of  a  Figure  is  its  position  in  a  number. 

A  figure  standing  alone,  or  in  the  first  place  at  the  right  of  other 
figures,  expresses  ones,  or  units  of  the  first  order. 

A  figure  in  the  second  place  expresses  tens,  or  units  of  the 
second  order. 

A  figure  in  the  third  place  expresses  hundreds,  or  units  of  the 
third  order ;  and  so  on. 

A  Period  is  a  group  of  three  orders  of  units,  counting  from  right 
to  left. 

Rule  for  Notation.  —  Begin  at  the  left,  and  write  the  hun- 
dreds, tens,  and  units  of  each  period  in  succession,  filling  vacant 
places  and  periods  with  ciphers. 


11  SUPPLEMENT 

Rule  for  Numeration.  —  Beginning  at  the  rights  separate  the 
number  into  periods. 

Beginning  at  the  left,  read  the  numbers  in  each  period,  giving 
the  narne  of  each  period  except  the  last. 

ADDITION 

Addition  is  finding  a  number  equal  to  two  or  more  given  num- 
bers. 
Addends  are  the  numbers  added. 
Tlie  Snm,  or  Amount,  is  the  number  obtained  by  addition. 

Principle.  —  Only  like  numbers,  and  units  of  the  same  order 
can  be  added. 

Rule.  —  Write  the  numbers  so  that  units  of  the  same  order  shall 
f  e  in  the  same  column. 

Beginning  at  the  right,  add  each  column  separately,  and  write 
the  su7n,  if  less  than  ten,  under  the  column  added. 

When  the  sum  of  any  column  exceeds  nine,  wHte  the  units  only, 
and  add  the  ten  or  tens  to  the  next  column. 

Write  the  entire  su7?i  of  the  last  column. 

SUBTRACTION 

Subtraction  is  finding  the  difference  between  two  numbers. 
The  Subtrahend  is  the  number  subtracted. 
The  Minuend  is  the  number  from  which  the  subtrahend  is  taken. 
The  Eemainder,  or  Difference,  is  the  number  left  after  subtracting 
one  number  from  another. 

Principles.  —  Only  like  numbers  and  units  of  the  same  order 
can  be  subtracted. 

The  sum  of  the  difference  and  the  subtrahend  inust  equal  the 
minuend. 

Rules.  —  I.  Write  the  subtrahend  under  the  minuend,  placing 
units  of  the  same  order  in  the  same  column. 


DEFINITIONS,    PRINCIPLES,    AND    RULES  111 

Beginning  at  the  right,  find  the  nuinber  that  7nust  be  added  to 
the  first  figure  of  the  subtrahend  to  produce  the  figure  in  the  corre- 
sponding order  of  the  minuend,  and  lurite  it  below.  Proceed  in 
this  way  until  the  difference  is  found. 

If  any  figure  in  the  subtrahend  is  greater  than  the  corresponding 
figure  in  the  minuend,  find  the  number  that  must  be  added  to  the 
former  to  produce  the  latter  increased  by  ten ;  then  add  one  to  the 
next  order  of  the  subtrahend  and  proceed  as  before. 

II.  Beginning  at  the  units  column,  subtract  each  figure,  of  the 
subtrahend  from  the  corresponding  figure  of  the  minuend  and 
write  the  remainder  below. 

If  any  figure  of  the  subtrahend  is  greater  than  the  corresponding 
figure  in  the  7ninuend,  add  ten  to  the  latter  and  subtract;  then, 
(a)  add  one  to  the  next  order  of  the  subtrahend  and  proceed  as 
before ;  or,  (b)  subtract  one  from  the  next  orden^  of  the  minuend 
and  proceed  as  before. 

MULTIPLICATION 

Multiplication  is  taking  one  number  as  many  times  as  there  are 
units  in  another  number. 

The  Multiplicand  is  the  number  taken  or  multiplied. 

The  Multiplier  is  the  number  that  shows  how  many  times  the 
multiplicand  is  taken. 

The  Product  is  the  result  obtained  by  multiplication. 

Principles.  —  The  multiplier  must  be  an  abstract  number. 
The  multiplicand  and  the  product  are  like  numbers. 
The  product  is  the  saine  in  whatever  order  the  numbers  are 
multiplied. 

Rule.  —  Write  the  m,ultiplier  under  the  multiplicand,  placing 
units  of  the  same  order  in  the  same  colutnn. 

Beginning  at  the  right,  Tnultiply  the  multiplicand  by  the  number 
of  units  in  each  order  of  the  multiplier  in  succession.      Write  the 


IV  SUPPLEMENT 

figure  of  the  lowest  order  in  each  partial  product  under  the  figure 
of  the  multiplier  that  produces  it.     Add  the  partial  products. 

To  multiply  by  10,  100,  1000,  etc. 

Rule.  —  Annex  as  many  ciphers  to  the  multiplicand  as  there 
are  cipher's  in  the  multiplier. 

DIVISION 

Division  is  finding  how  many  times  one  number  is  contained  in 
another,  or  finding  one  of  the  equal  parts  of  a  number. 

The  Dividend  is  the  number  divided. 
.  The  Divisor  is  the  number  contained  in  the  dividend. 
The  Quotient  is  the  result  obtained  by  division. 

Principles.  —  When  the  divisor  and  the  dividend  are  like  num- 
bers, the  quotient  is  an  abstract  number. 

When  the  divisor  is  an  abstract  number^  the  dividend  and  the 
quotient  are  like  numbe)'S. 

The  product  of  the  divisor  and  the  quotient,  plus  the  remainder, 
if  any,  is  equal  to  the  dividend. 

Rule.  —  Write  the  divisor  at  the  left  of  the  dividend  with  a  line 
between  them. 

Find  how  many  times  the  divisor  is  contained  in  the  fewest  fig- 
ures on  the  left  of  the  dividend,  and  write  the  result  over  the  last 
figure  of  the  partial  dividend.  Multiply  the  divisor  by  this  quotient 
figure,  and  write  the  product  under  the  figures  divided.  Subtract 
the  product  from  the  partial  dividend  used,  and  to  the  remainder 
annex  the  next  figure  of  the  dividend  for  a  new  dividend. 

Divide  as  before  until  all  the  figures  of  the  dividend  have  been 
used. 

If  any  partial  dividend  will  not  contain  the  divisor,  write  a 
ciphei'  in  the  quotient,  and  annex  the  next  figure  of  the  dividend. 

If  there  is  a  remainder  aflei'  the  last  division,  write  it  after  the 
quotient  with  the  divisor  underneath. 


DEFINITIONS,    PRINCIPLES,    AND   RULES  V 

FACTORING 

An  Exact  Divisor  of  a  number  is  a  number  that  will  divide  it 
without  a  remainder. 

An  Odd  Number  is  one  that  cannot  be  exactly  divided  by  two. 

An  Even  Number  is  one  that  can  be  exactly  divided  by  two. 

The  Factors  of  a  number  are  the  numbers  that  multiplied  to- 
gether produce  that  number. 

A  Prime  Number  is  a  number  that  has  no  factors. 

A  Composite  Number  is  a  number  that  has  factors. 

A  Prime  Pactor  is  a  prime  number  used  as  a  factor. 

A  Composite  Factor  is  a  composite  number  used  as  a  factor. 

Factoring  is  separating  a  number  into  its  factors. 

To  find  the  Prime  Factors  of  a  Number. 

Rule.  —  Divide  the  number  by  any  prime  factor.  Divide  the 
quotient,  if  composite,  in  like  m^anner ;  and  so  continue  until  a 
prime  quotient  is  found.  The  several  divisors  and  the  last  quotient 
will  be  the  prime  factors. 

CANCELLATION 

Cancellation  is  rejecting  equal  factors  from  dividend  and  divisor. 
Principle.  —  Dividing  dividend  and  divisor  by  the  same  num^ 
ber  does  not  affect  the  quotient. 

GREATEST  COMMON  DIVISOR 

A  Common  Factor  (divisor  or  measure)  is  a  number  that  is  a 
factor  of  each  of  two  or  more  numbers. 

A  Common  Prime  Factor  is  a  prime  number  that  is  a  factor  of 
each  of  two  or  more  numbers. 

The  Greatest  Common  Factor  (divisor  or  measure)  is  the  largest 
number  that  is  a  factor  of  each  of  two  or  more  numbers. 

Numbers  are  prime  to  each  other  when  they  have  no  common 
factor. 


VI  SUPPLEMENT 

The  greatest  common  divisor  of  two  or  more  numbers  is  the 
product  of  their  common  prime  factors. 

PRiNCiPiiES.  —  A  common  divisor  of  two  numbers  is  a  divisor 
of  their  sum,  and  also  of  their  difference. 

A  divisor  of  a  number  is  a  divisor  of  eve^y  multiple  of  that 
nu77iber ;  and  a  common  divism'  of  two  or  more  number's  is  a 
divisor  of  any  of  their  multiples. 

To  find  the  Common  Prime  Factors  of  Two  or  More  Numbers. 

Rule,  —  Divide  the  numbers  by  any  common  prime  factors, 
and  the  quotients  in  like  manner,  until  they  have  no  common 
factor;  the  several  divisors  are  the  common  prime  factors. 

To  find  the  Greatest  Oommon  Divisor  of  Numbers  that  are  Easily 
Factored. 

Rule.  —  Separate  the  numbers  into  thdr  prime  factors ;  the 
product  of  those  that  are  common  is  the  greatest  common  divisor. 

To  find  the  Greatest  Oommon  Divisor  of  Numbers  that  are  not 
Easily  Factored. 

Rule.  —  Divide  the  greater  number  by  the  less;  then  divide 
the  last  divisor  by  the  last  remainder,  continuing  until  there  is  no 
remainder.     The  last  divisor  is  the  greatest  common  divisoi\ 

If  there  are  more  than  two  numbers,  find  the  greatest  common 
divisor  of  two  of  theyn;  then  of  that  divisor  and  another  of  the 
numbers  until  all  of  the  nu7nbers  have  been  used.  The  last  divisor 
is  the  greatest  common  divisor. 

LEAST  COMMON  MULTIPLE 

A  Multiple  of  a  number  is  a  number  that  exactly  contains  that 
number. 

A  Oommon  Multiple  of  two  or  more  numbers  is  a  number  that 
is  a  multiple  of  each  of  them. 

The  Least  Oommon  Multiple  of  two  or  more  numbers  is  the 
smallest  number  that  is  a  common  multiple  of  them. 


DEFINITIONS,    PRINCIPLE^^MS^IlfiSj^iTY  j    y^ 

Principles.  —  A  multiple  of  a  number  comSM^IflTthe  prime 
factors  of  that  number. 

A  cortimon  multiple  of  two  or  m,ore  numbers  contains  each  of 
the  prime  factors  of  those  numbers. 

The  Least  Common  Multiple  of  tivo  or  more  numbers  contains 
only  the  prime  factors  of  each  of  the  numbers. 

To  find  the  Least  Common  Multiple  of  Two  or  More  Numbers. 

Rule.  —  Divide  by  any  prime  numbei'  that  is  an  exact  divisor  of 
two  or  more  of  the  numbers,  and  write  the  quotients  and  undivided 
numbers  below.  Divide  these  numbers  in  like  manner,  contimiing 
until  no  two  of  the  remaining  numbers  have  a  common  factor. 
The  product  of  the  divisors  and  remaining  numbers  is  the  least 
common  multiple. 

FRACTIONS 

A  Fraction  is  one  or  more  of  the  equal  parts  of  anything. 

The  Unit  of  a  Praction  is  the  number  or  thing  that  is  divided 
into  equal  parts. 

A  Fractional  Unit  is  one  of  the  equal  parts  into  which  the  num- 
ber or  thing  is  divided. 

The  Terms  of  a  Fraction  are  its  numerator  and  its  denominator. 

The  Denominator  of  a  fraction  shows  into  how  many  parts  the 
unit  is  divided. 

The  Numerator  of  a  fraction  shows  how  many  of  the  parts  are 
taken. 

A  fraction  indicates  division ;  the  numerator  being  the  divi- 
dend and  the  denominator  the  divisor. 

The  Value  of  a  Fraction  is  the  quotient  of  the  numerator  divided 
by  the  denominator. 

Fractions  are  divided  into  two  classes  —  Common  and  Decimal. 

A  Common  Fraction  is  one  in  which  the  unit  is  divided  into  any 
number  of  equal  parts. 

A  common  fraction  is  expressed  by  writing  the  numerator  above 
the  denominator  with  a  dividing  line  between. 


VIU  SUPPLEMENT 

Common  fractions  consist  of  three  principal  classes  —  Simple, 
Compound,  and  Complex. 

A  Simple  Fraction  is  one  whose  terms  are  whole  numbers. 

A  Proper  Fraction  is  a  simple  fraction  whose  numerator  is  less 
than  its  denominator. 

An  Improper  Fraction  is  a  simple  fraction  whose  numerator 
equals  or  exceeds  its  denominator. 

A  Compound  Fraction  is  a  fraction  of  a  fraction. 

A  Complex  Fraction  is  one  having  a  fraction  in  its  numerator,  or 
in  its  denominator,  or  in  both. 

A  Mixed  Number  is  a  whole  number  and  a  fraction  written 
together. 

The  Keciprocal  of  a  Number  is  one  divided  by  that  number. 

The  Eeciprocal  of  a  Fraction  is  one  divided  by  the  fraction,  or 
the  fraction  inverted. 

Principles.  —  Multiplying  the  numerator  or  dividing  the  de- 
nominator multiplies  the  fraction. 

Dividing  the  numerator  or  multiplying  the  denominator  divides 
the  fraction. 

Multiplying  or  dividing  both  terms  of  a  fraction  by  the  same 
number  does  not  alter  the  value  of  the  fraction. 

Eeduction  of  fractions  is  changing  their  terms  without  altering 
their  value. 

To  reduce  a  Fraction  to  Higher  Terms. 

Rule.  —  Multiply  both  numerator  and  denominator  by  the  same 
number. 

To  reduce  a  Fraction  to  its  Lowest  Terms. 

Rule.  —  Divide  both  te^-ms  of  the  fraction  by  their  greatest 
common  divisor. 

A  fraction  is  in  its  lowest  terms  wIkmi  the  numerator  and  the 
denominator  are  prime  to  each  other. 


DEFINITIONS,    PRINCIPLES,    AND   RULES  ix 

To  reduce  a  Mixed  Number  to  an  Improper  Fraction. 

Rule.  —  Multiply  the  whole  number  by  the  denominator ;  to  the 
product  add  the  numerator ;  and  place  the  sum  over  the  denom- 
inator. 

To  reduce  an  Improper  Fraction  to  a  Whole  or  to  a  Mixed  Number. 

Rule.  —  Divide  the  num^erator  by  the  denominator. 

A  Common  Denominator  is  a  denominator  common  to  two  or 
more  fractions. 

The  Least  Common  Denominator  is  the  smallest  denominator 
common  to  two  or  more  fractions. 

To  reduce  Fractions  to  their  Least  Common  Denominator, 

Rule.  —  Find  the  least  common  viultiple  of  all  the  denomi- 
nators/or the  least  common  denominator.  Divide  this  multiple  by 
the  denominator  of  each  fraction^  and  Tnultiply  the  numerator  by 
the  quotient. 

ADDITION  OF  FRACTIONS 

Principle.  —  Only  like  fractions  can  be  added. 

Rule.  — Reduce  the  fractions,  if  necessary,  to  a  comm,on  denom- 
inator, and  over  it  write  the  sum,  of  the  numerators. 

If  there  are  mixed  numbers,  add  the  fractions  and  the  whole 
numbers  separately ^  and  unite  the  results. 

SUBTRACTION  OF  FRACTIONS 

Principle.  —  Only  like  fractions  can  be  subtracted. 

Rule.  —  Reduce  the  fractions,  if  necessary,  to  a  common  denom- 
inator, and  over  it  write  the  difference  between  the  numerators. 

If  there  are  mixed  numbers  subtract  the  fractions  and  the  whole 
numbers  separately,  and  unite  the  results. 

MULTIPLICATION  OF   FRACTIONS 

Rule.  —  Reduce  whole  and  mixed  numbers  to  improper  frac- 
tions ;  cancel  the  factors  common  to  numerators  and  denomina- 
tors, and  write  the  product  of  the  remaining  factors  in  the  nitmer- 
ators  over  the  product  of  the  remaining  factors  in  the  denominators. 


SUPPLEMENT 


DIVISION   OF  FRACTIONS 


Rules.  —  I.  Reduce  whole  and  mixed  numbers  to  improper 
fractions.  Reduce  the  fractions  to  a  common  denominator.  Divide 
the  numerator  of  the  dividend  hy  the  nume^^ator  of  the  divisor. 

II.  Invert  the  divisor  and  proceed  as  in  multiplication  of  frac- 
tions. 

To  reduce  a  Oomplex  Fraction  to  a  Simple  One. 

Rules.  —  I.  Multiply  the  numerator  of  the  complex  fraction 
by  its  denoTninator  inverted. 

II.  Multiply  both  terms  by  the  least  common  multiple  of  the 
denominators. 

DECIMALS 

A  Decimal  Fraction  is  one  in  which  the  unit  is  divided  into 
tenths,  hundredths,  thousandths,  etc. 

A  Decimal  is  a  decimal  fraction  whose  denomination  is  indi- 
cated by  the  number  of  places  at  the  right  of  the  decimal  point. 

The  Decimal  Point  is  the  mark  used  to  locate  units. 

A  Mixed  Decimal  is  a  whole  number  and  a  decimal  written 
together. 

A  Oomplex  Decimal  is  a  decimal  with  a  common  fraction 
written  at  its  right. 

To  write  Decimals. 

Rule.  —  Write  the  numerator ;  and  from  the  right,  point  off  as 
many  decimal  places  as  there  are  ciphers  in  the  denominator, 
prefixing  ciphers,  if  necessary,  to  make  the  required  nu/mber. 

To  read  Decimals. 

Rule.  —  Read  the  numerator,  and  give  the  name  of  the  right- 
hand  order. 

Principles.  —  Prefixing  ciphers  to  a  decimal  diminishes  its 
valv£. 


DEFINITIONS,    PRINCIPLES,    AND   RULES  xi 

Removing  ciphers  from  the  left  of  a  decimal  increases  its  value. 
Annexing  ciphers  to  a  decimal  or  removing  ciphers  fronn  its 
right  does  not  alter  its  value. 

To  reduce  a  Decimal  to  a  Common  Fraction. 

Rule. —  Write  the  figures  of  the  decimal  for  the  numerator,  and 
1,  with  as  many  ciphers  as  there  are  places  in  the  decimal,  for  the 
denominator,  and  reduce  the  fraction  to  its  lowest  terms. 

To  reduce  a  Common  Fraction  to  a  Decimal. 

Rule.  —  Annex  decimal  ciphers  to  the  numerator,  and  divide  it 
by  the  denominator. 

To  reduce  Decimals  to  a  Common  Denominator, 

Rule.  —  Make  their  decimal  places  equal  by  annexing  ciphers. 

ADDITION  AND  SUBTRACTION  OF  DECIMALS 
Decimals  are  added  and  subtracted  the  same  as  whole  numbers. 

MULTIPLICATION  OF  DECIMALS 

Rule.  —  Multiply  as  in  whole  numbers,  and  from  the  right  of 
the  product,  point  off  as  many  decimal  places  as  there  are  decimal 
places  in  both  factors. 

DIVISION  OF  DECIMALS 

Rule.  —  Make  the  divisor  a  whole  number  by  removing  the 
decimal  point,  and  make  a  corresponding  change  in  the  dividend. 
Divide  as  in  whole  numbers,  and  place  the  decimal  point  in  the 
quotient  under  {or  over)  the  new  decimal  point  in  the  dividend. 

ACCOUNTS  AND  BILLS 

A  Debtor  is  a  person  who  owes  another. 

A  Creditor  is  a  person  to  whom  a  debt  is  due. 


Xll  SUPPLEMENT 

An  Account  is  a  record  of  debits  and  credits  between  persons 
doing  business. 

The  Balance  of  an  account  is  the  diflference  between  the  debit 
and  credit  sides. 

A  Bill  is  a  written  statement  of  an  account. 

An  Invoice  is  a  written  statement  of  items,  sent  with  merchan- 
dise. 

A  Beceipt  is  a  written  acknowledgment  of  the  payment  of 
part  or  all  of  a  debt. 

A  bill  is  receipted  when  the  words,  "  Received  Payment,"  are 
written  at  the  bottom,  signed  by  the  creditor,  or  by  some  person 
duly  authorized. 

DENOMINATE  NUMBERS 

A  Measure  is  a  standard  established  by  law  or  custom,  by 
which  distance,  capacity,  surface,  time,  or  weight  is  determined. 

A  Denominate  Unit  is  a  unit  of  measure. 

A  Denominate  Number  is  a  denominate  unit  or  a  collection  of 
denominate  units. 

A  Simple  Denominate  Number  consists  of  denominate  units  of 
one  kind. 

A  Compound  Denominate  Number  consists  of  denominate  units  of 
two  or  more  kinds. 

A  Denominate  Fraction  is  a  fraction  of  a  denominate  number. 

A  denominate  fraction  may  be  either  common  or  decimal, 

Eeduction  of  denominate  numbers  is  changing  them  from  one 
denomination  to  another  without  altering  their  value. 

Eeduction  Descending  is  changing  a  denominate  number  to  one 
of  a  lower  denomination.  -' 

Rule.  —  Multiply  the  highest  denomination  hy  the  number  re- 
quired to  reduce  it  to  the  next  lowei'  dctioTnination,  and  to  the  prod- 
uct add  the  units  of  thai  lower  denomination,  if  any.  Proceed 
in  this  manner  until  the  required  denomination  is  reached. 


DEFINITIONS,    PKINCIPLES,    AND    RULES  xiii 

Eednction  Ascending  is  changing  a  denominate  number  to  one  of 
a  higher  denomination. 

Rule.  —  Divide  the  given  denomination  successively  hy  the 
numbers  that  will  reduce  it  to  the  required  deno7nination.  To  this 
quotient  annex  the  several  remainders. 

To  find  the  Time  between  Dates. 

Rule.  —  When  the  time  is  less  than  one  year,  find  the  exact 
number  of  days;  if  greate)'  than  one  year,  find  the  time  by  com- 
pound subtraction,  talcing  30  days  to  the  month. 

PERCENTAGE 

Per  Oent  means  hundredths. 

Percentage  is  computing  by  hundredths. 

The  elements  involved  in  percentage  are  the  Base,  Bate,  Per- 
centage, Amount,  and  Difference. 

The  Base  is  the  number  of  which  a  number  of  hundredths  is 
taken. 

The  Kate  indicates  the  number  of  hundredths  to  be  taken. 

The  Percentage  is  one  or  more  hundredths  of  the  base. 

The  Amount  is  the  base  increased  by  the  percentage. 

The  Difference  is  the  base  diminished  by  the  percentage. 

To  find  the  Percentage  when  the  Base  and  Eate  are  Given. 
Rule.  —  Multiply  the  base  by  the  rate  expressed  as  hundredths. 
To  find  the  Eate  when  the  Percentage  and  Base  are  Given. 
Rule.  —  Divide  the  percentage  by  the  base. 
To  find  the  Base  when  the  Percentage  and  Eate  are  Given. 
Rule.  —  Divide  the  percentage  by  the  rate  expressed  as  hun- 
dredths. 

To  find  the  Base  when  the  Amount  and  Eate  are  Given. 
Rule.  —  Divide  the  amount  by  l-{-  the  rate  expressed  as  hun- 
dredths. 


XIV  SUPPLEMENT 

To  find  the  Base  when  the  Difference  and  Eate  are  Given. 

Rule.  —  Divide  the  diff cogence  by  \-~the  rate  expressed  as  hun- 
dredths. 

PROFIT  AND   LOSS 

Profit  or  Loss  is  the  difference  between  the  buying  and  selling 
prices. 

In  Profit  and  Loss, 

The  buying  price,  or  cost,  is  the  base. 
The  rate  per  cent  profit  or  loss  is  the  rate. 
The  profit  or  loss  is  the  percentage. 

The  selling  price  is  the  amount  or  difference,  according  as  it 
is  more  or  less  than  the  buying  price. 

COMMERCIAL  DISCOUNT 

Oonuneroial  Discount  is  a  percentage  deducted  from  the  list 
price  of  goods,  the  face  of  a  bill,  etc. 

The  Net  Price  of  goods  is  the  sum  received  for  them. 

In  Oommercial  Discount, 

The  list  price,  or       )  .     •,     7 

The  face  of  the  bill  |  ^^  the  Josa. 

The  rate  per  cent  discount  is  the  rate. 

The  discount  is  the  percentage. 

The  list  price  diminished  by  the  discount  is  the  difference. 

In  successive  discounts,  the  first  discount  is  made  from  the  list 
price  or  the  face  of  the  bill ;  the  second  discount,  from  the  list 
price  or  face  of  the  bill  diminished  by  the  first  discount ;  and  so 
on. 

COMMISSION 

Commission  is  a  percentage  allowed  an  agent  for  his  services. 
A  Oommission  Agent  is  one  who  transacts  business  on   com- 
mission. 


DEFINITIONS,    PRINCIPLES,    AND    RULES  XV 

A  Oonsignment  is  the  merchandise  forwarded  to  a  commission 
agent. 

The  Consignor  is  the  person  who  sends  the  merchandise. 

The  Consignee  is  the  person  to  whom  the  merchandise  is  sent. 

The  Net  Proceeds  is  the  sum  remaining  after  all  charges  have 
been  deducted. 

In  buying,  the  commission  is  a  percentage  of  the  buying  price; 
in  selling,  a  percentage  of  the  selling  price;  in  collecting,  a  per- 
centage of  the  sunfi  collected;   hence  : 

The  sum  invested,  or 


>-  is  the  base. 
The  sum  collected         ) 


The  rate  per  cent  commission  is  the  rate. 
The  commission  is  the  percentage. 

The  sum  invested  increased  by  the  commission  is  the  amount. 
The  sum  collected  diminished  by  the  commission  is  the  differ- 
ence. 

INSURANCE 

Insurance  is  a  contract  of  indemnity. 

Insurance  is  of  three  kinds  —  Fire,  Marine,  and  Life. 

Fire  Insurance  is  indemnity  against  loss  of  property  by  fire. 

Marine  Insurance  is  indemnity  against  loss  of  property  by  the 
casualities  of  navigation. 

Life  Insurance  is  indemnity  against  loss  of  life. 

The  Insurance  Policy  is  the  contract  setting  forth  the  liability  of 
the  insurer. 

The  Pohcy  Face  is  the  amount  of  insurance. 

The  Premium  is  the  price  paid  for  insurance. 

The  Insurer,  or  Underwriter,  is  the  company  issuing  the  policy. 

The  Insured  is  the  person  for  whose  benefit  the  policy  is  issued. 

In  Insurance, 

The  policy  face  is  the  base. 

The  rate  per  cent  premium  is  the  rate. 

The  premium  is  the  percentage. 


XVi  SUPPLEMENT 

TAXES 

A  Tax  is  a  sum  of  money  levied  on  persons  or  property  foi 
public  purposes. 

A  Personal,  or  Poll  Tax,  is  a  tax  on  the  person. 

A  Property  Tax  is  a  tax  of  a  certain  per  cent  on  the  assessed 
value  of  property. 

Property  may  be  either  personal  or  real. 

Personal  Property  consists  of  such  things  as  are  movable. 

Keal  Property  is  that  which  is  fixed,  or  immovable. 

In  Taxes, 

The  assessed  value  is  the  base. 
The  rate  of  taxation  is  the  rate. 
The  tax  is  the  percentage. 

DUTIES 

Duties  are  taxes  on  imported  goods. 
Duties  are  either  Specific  or  Ad  Valorem. 
A  Specific  Duty  is  a  tax  on  goods  without  regard  to  cost. 
An  Ad  Valorem  duty  is  a  tax  of  a  certain  per  cent  on  the  cost 
of  goods. 

In  Ad  Valorem  Duties, 
The  cost  of  the  goods  is  the  base. 
The  rate  per  cent  duty  is  the  rate. 
The  ad  valorem  duty  is  the  percentage. 

INTEREST 

Interest  is  the  sum  paid  for  the  use  of  money. 

Tlie  Principal  is  the  sum  loaned. 

The  Amount  is  the  sum  of  the  principal  and  interest. 

The  Eate  of  Interest  is  the  rate  per  cent  for  one  year. 

The  Legal  Rate  is  the  rate  fixed  by  law. 

Usury  is  interest  at  a  higher  rate  than  that  fixed  by  law. 

Simple  Interest  is  interest  on  the  principal  only. 


DEFINITIONS,    PRINCIPLES,    AND   RULES  Xvii 

To  find  the  Interest  when  the  Principal,  Time,  and  Eate  are  Given. 

KuLE.  —  Multiply  the  principal  by  the  rate  expressed  as  hun- 
dredths, and  this  product  by  the  time  expressed  in  years. 

To  find  the  Time  when  the  Principal,  Interest,  and  Rate  are  Given. 

Rule.  —  Divide  the  given  interest  by  the  interest  for  one  year. 

To  find  the  Eate  when  the  Principal,  Interest,  and  Time  are  Given. 

Rule.  —  Divide  the  given  interest  by  the  interest  at  one  per 
cent. 

To  find  the  Principal  when  the  Interest,  Eate,  and  Time  are  Given. 
Rule.  —  Divide  the  given  interest  by  the  interest  on  $  1. 

To  find  the  Principal  when  the  Amount  and  Time  and  Eate  are 
Given. 

Rule.  —  Divide  the  given  amount  by  the  a7nount  of  $1. 

Interest  by  Aliquot  Parts. 

To  find  the  Interest  for  Years,  Months,  and  Days. 

Rule.  —  Mnd  the  interest  for  one  year  and  take  ihis  as  many 
times  as  there  are  years. 

Take  the  greatest  number  of  the  given  months  that  equals  an 
aliquot  part  of  a  year  and  find  the  interest  for  this  time.  Take 
aliquot  parts  of  this  for  the  remaining  months. 

In  the  same  manner  find  the  interest  for  the  days. 

The  sum  of  these  interests  will  be  the  interest  required. 

To  find  the  Interest  when  the  Time  is  Less  than  a  Tear. 

Rule. —  Find  the  interest  for  the  time  in  months  or  days  that 
will  gain  one  per  cent  of  the  principal. 

Find  by  aliquot  parts,  as  in  the  first  rule,  the  interest  for  the 
remaining  time. 

The  sum  of  these  interests  will  be  the  int&rest  required. 


XVlll  SUPPLEMENT 

Interest  by  Six  Per  Cent  Method. 

To  find  the  Interest  at  6%. 

Rule.  —  For  Years:  Multiply  the  principal  hy  the  rate  ex- 
pressed as  hundredths f  and  that  product  hy  the  number  of  years. 

For  Months :  Move  the  decimal  point  two  places  to  the  left^  and 
multiply  by  one-half  the  number  of  months. 

For  Days !  Move  the  decimal  point  three  places  to  the  left,  and 
multiply  by  one-sixth  the  number  of  days. 

To  find  the  interest  at  any  other  rate  per  cent,  divide  the  in- 
terest at  6%  by  6,  and  multiply  the  quotient  by  the  given  rate. 

To  find  Exact  Interest. 

Rule.  —  Multiply  the  principal  by  the  rate  expressed  as  hun- 
dredths, and  thai  product  by  the  time  expressed  in  years  of  365 
days. 

ANNUAL  INTEREST 

Annual  Interest  is  interest  payable  annually.  If  not  paid  when 
due,  annual  interest  draws  simple  interest. 

To  find  the  Amount  Due  on  a  Note  with  Annual  Interest,  when  the 
Interest  has  not  been  Paid  Annually. 

Rule.  —  Find  the  interest  on  the  principal  for  the  entire  time, 
and  on  each  annual  interest  for  the  time  it  remained  unpaid. 
The  sum  of  the  principal  and  all  the  interest  is  the  amount  due. 

COMPOUND   INTEREST 

Oompound  Interest  is  interest  on  the  principal  and  on  the  un- 
paid interest,  which  is  added  to  the  principal  at  regular  inter- 
nals. The  interest  may  be  compounded  annually,  semi-annually, 
quarterly,  etc.,  according  to  agreement. 

To  find  Oompound  Interest. 

Rule.  —  Find  the  amount  of  the  given  principal  for  the  first 
periods     Considering  this  as  a  ncio  principal^  find  the  amount  of 


DEFINITIONS,    PRINCIPLES,    AND    RULES  xix 

it  for  the  next  period,  continuing  in  this  manner  for  the  given 
time. 

Find  the  difference  between  the  last  amount  and  the  given 
principal,  which  will  be  the  compound  interest. 

PARTIAL   PAYMENTS 

Partial  Payments  are  part  payments  of  a  note  or  debt.  Eacli 
payment  is  recorded  on  the  back  of  the  note  or  the  written 
obligation. 

United  States  Rule.  —  Find  the  amount  of  the  principal  to 
the  time  when  the  payment  or  the  sum  of  two  or  more  payments 
equals  or  exceeds  the  interest. 

From  this  amount  deduct  the  payment  or  sum  of  payments. 

Use  the  balance  then  due  as  a  new  principal,  and  proceed  as 
before. 

Merchants'  Rule.  —  Find  the  amount  of  an  interest-bearing 
note  at  the  time  of  settlement. 

Fi7id  the  amount  of  each  credit  from  its  time  of  payment  to  the 
time  of  settleftnent ;  subtract  their  sum  from  the  amount  of  the 
principal. 

BANK  DISCOUNT 

Bank  Discount  is  a  percentage  retained  by  a  bank  for  advanc- 
ing money  on  a  note  before  it  is  due. 

The  Sum  Discounted  is  the  face  of  the  note,  or  if  interest-bear- 
ing, the  amount  of  the  note  at  maturity. 

The  Term  of  Discount  is  the  number  of  days  from  the  day  of 
discount  to  the  day  of  maturity. 

The  Bank  Discount  is  the  interest  on  the  sum  discounted  for 
the  term  of  discount. 

The  Proceeds  of  a  note  is  the  sum  discounted  less  the  bank  dis- 
count. 

Problems  in  bank  discount  are  calculated  as  problems  in 
interest. 


XX  SUPPLEMENT 

In  Bank  Discount, 

The  sum  discounted  is  the  priiicipal. 

The  rate  of  discount  is  the  rate  of  interest. 

The  term  of  discount  is  the  time. 

The  bank  discount  is  the  proceeds. 

EXCHANGE 

Exchange  is  making  payments  at  a  distance  by  means  of  drafts 
or  bills  of  exchange. 

Domestic  Exchange  is  exchange  between  places  in  the  same 
country. 

Foreign  Exchange  is  exchange  between  different  countries. 

Exchange  is  at  par  when  a  draft,  or  bill,  sells  for  its  face 
value ;  at  a  premium  when  it  sells  for  more  than  its  face  value  ; 
at  a  discount  when  it  sells  for  less. 

The  cost  of  a  sight  draft  is  the  face  of  the  draft  increased  by 
the  premium,  or  diminished  by  the  discount. 

The  cost  of  a  time  draft  is  the  face  of  the  draft  increased  by 
the  premium,  or  diminished  by  the  discount,  and  this  result 
diminished  by  the  bank  discount. 

To  find  the  Oost  of  a  Draft. 

Rule. ^-jpmc?  the  cost  of  $1  of  the  draft;  multiply  this  hy  the 
face  of  the  draft. 

To  find  the  Pace  of  a  Draft. 

Rule.  —  Divide  the  cost  of  the  draft  hy  the  cost  q/"  $  1  of  the 
draft. 

EQUATION  OF  PAYMENTS 

Equation  of  Payments  is  a  method  of  ascertaining  at  what  time 
several  debts  due  at  different  times  may  be  settled  by  a  single 
payment. 

The  Equated  Time  of  payment  is  the  time  when  the  several 
debts  may  be  equitably  settled  by  one  payment. 

The  Term  of  Credit  is  the  time  the  debt  has  to  run  before  it 
becomes  due. 


DEFINITIONS,    PRINCIPLES,    AND   RULES  xxi 

The  Average  Term  of  Credit  is  the  time  the  debts  due  at  different 
times  have  to  run,  before  they  may  be  equitably  settled  by  one 
payment. 

To  find  the  Equated  Time  of  Payment  when  the  Terms  of  Credit 
begin  at  the  Same  Date. 

Rule.  —  Multiply  each  debt  by  its  term  of  credit,  and  divide  the 
sum  of  the  products  by  the  sum  of  the  debts.  The  quotient  will  be 
the  average  term  of  credit. 

Add  the  average  term,  of  credit  to  the  date  of  the  debts,  and  the 
result  will  be  the  equated  time  of  payment. 

To  find  the-  Equated  Time  when  the  Terms  of  Credit  begin  at 
Different  Dates. 

Rule.  — Find  the  date  at  which  each  debt  becomes  due.  Select 
the  earliest  date  as  a  standard. 

Multiply  each  debt  by  the  number  of  days  between  the  standard 
date  and  the  date  whe7i  the  debt  becomes  due,  and  divide  the  sum 
of  the  products  by  the  sum  of  the  debts.  The  quotient  will  be  the 
average  term  of  credit  from  the  standard  date. 

Add  the  average  term  of  credit  to  the  standard  date,  and  the 
result  will  be  the  equated  time  of  payment, 

RATIO 

Batio  is  the  relation  one  number  bears  to  another  of  the  same 
kind. 

The  Terms  of  the  ratio  are  the  numbers  compared. 

The  Antecedent  is  the  first  term. 

The  Consequent  is  the  second  term. 

The  antecedent  and  consequent  form  a  couplet. 

Principles.  —  See  Fractions. 

PROPORTION 

A  Proportion  is  formed  by  two  equal  ratios. 

The  Extremes  of  a  proportion  are  the  first  and  last  terms. 

The  Means  of  a  proportion  are  the  second  and  third  terms. 


XXU  SUPPLEMENT 

Principles.  —  The  ^product  of  the  means  is  equal  to  the  prod- 
uct of  the  extremes. 

Either  mean  equals  the  product  of  the  extremes  divided  by  the 
other  m,ean. 

Either  extreme  equals  the  product  of  the  means  divided  hy  the 
other  extreme. 

Rule  for  Proportion.  —  Represent  the  required  term  by  x. 

Arrange  the  terms  so  that  the  required  tei^m  and  the  similar 
known  term  may  form  one  couplet,  the  remaining  tet-ms  the  other. 

If  the  required  term  is  in  the  extremes,  divide  the  product  of  the 
means  by  the  given  extrefme. 

If  the  required  term  is  in  the  mean^,  divide  the  product  of  the 
extremes  by  the  given  mean. 

PARTNERSHIP 

Partnership  is  an  association  of  two  or  more  persons  for  busi- 
ness purposes. 

The  Partners  are  the  persons  associated. 

The  Capital  is  that  which  is  invested  in  the  business. 

The  Assets  are  the  partnership  property. 

The  Liabilities  are  the  partnership  debts. 

To  find  the  Profit,  or  Loss,  of  Each  Partner  when  the  Capital  of 
Each  is  Employed  for  the  Same  Period  of  Time. 

Rule.  —  Find  the  part  of  the  entire  profit,  or  bss,  that  each 
partner's  capital  is  of  the  entire  capital. 

To  find  the  Profit,  or  Loss,  of  Each  Partner  when  the  Capital  of 
Each  is  Employed  for  Different  Periods  of  Time. 

Rule.  —  Find  each  partner's  capital  for  one  month,  by  multi- 
plying the  amount  he  invests  by  the  number  of  months  it  is 
employed;  then  find  the  part  of  the  entire  profit,  or  loss,  that  each 
partner's  capital  for  one  month  is  of  the  entire  capital  for  one 
month. 


DEFINITIONS,    PEINCIPLES,    AND    RULES  Xxiii 

INVOLUTION 

A  Power  of  a  number  is  the  product  obtained  by  using  that 
number  a  certain  number  of  times  as  a  factor. 

The  First  Power  of  a  number  is  the  number  itself. 

The  Second  Power  of  a  number,  or  the  Square,  is  the  product  of 
a  number  taken  twice  as  a  factor. 

The  Third  Power  of  a  number,  or  the  Cube,  is  the  product  of  a 
number  taken  three  times  as  a  factor. 

An  Exponent  is  a  small  figure  written  a  little  to  the  right  of  the 
upper  part  of  a  number  to  indicate  the  power. 

Involution  is  finding  any  power  of  a  number. 

To  find  the  Power  of  a  Number. 

Rule. —  Take  the  number  as  a  factor-  as  Triany  times  as  there 
are  units  in  the  exponent. 

EVOLUTION 

A  Root  is  one  of  the  equal  factors  of  a  number. 

The  Square  Eoot  of  a  number  is  one  of  its  two  equal  factors. 

The  Cube  Eoot  of  a  number  is  one  of  its  three  equal  factors. 

Evolution  is  finding  any  root  of  a  number. 

Evolution  may  be  indicated  in  two  ways :  by  the  Radical 
Sign,  V~,  or  by  sl  fractional  exponent. 

The  Index  of  a  root  is  a  small  figure  placed  a  little  to  the  left 
of  the  upper  part  of  the  radical  sign,  to  indicate  what  root  is  to 
be  found.     In  expressing  square  root,  the  index  is  omitted. 

In  the  fractional  exponent,  the  numerator  indicates  the  power 
to  which  the  number  is  to  be  raised ;  the  denominator  indicates 
the  root  to  be  taken  of  the  number  thus  raised. 

To  find  the  Square  Root  of  a  Number. 

Rule.  —  Point  off  in  periods  of  two  figures,  commencing  at 
units.  Find  the  greatest  square  in  the  first  period  and  place  the 
root  in  the  quotient.  Subtract  this  square  from  the  first  period, 
and  bring  down  the  next  period. 


XXiV  SUPPLEMENT 

Multiply  the  quotient  figure  hy  two,  and  use  it  as  a  trial  divisor. 
Place  the  second  figure  in  the  quotient,  and  annex  it  also  to  the 
trial  divisor.  Then  multiply  the  figures  in  the  tiial  divisor  by  the 
second  quotient  figure,  and  subtract. 

Bring  down  the  next  period,  and  proceed  as  before  until  the 
square  root  is  found. 

To  find  the  Square  Eoot  of  a  Praction. 

Rule.  —  Reduce  the  fraction  to  its  simplest  form,  and  find  the 
square  root  of  each  term  separately. 

To  find  the  Onbe  Boot  of  a  Number. 

Rule.  —  Point  off  in  periods  of  three  figures  each,  beginning  at 
units. 

Find  the  greatest  cube  in  the  first  period  and  place  the  root  in 
the  quotient.  Subtract  this  cube  from  the  first  period,  and  bring 
down  the  next  period. 

Multiply  the  squxire  of  the  first  quotient  figure  by  three  and 
annex  two  ciphers  for  a  trial  divisor.  Place  the  second  figure  in 
the  quotient.  Then,  to  the  trial  divisor  add  three  times  the  prod- 
uct of  the  first  and  second  figures,  also  the  square  of  the  second. 
Multiply  this  sum  by  the  second  figure  and  subtract. 

Bring  down  the  next  period,  and  proceed  as  before  until  the  cube 
root  is  found. 

To  find  the  Oube  Eoot  of  a  Fraction. 

Rule.  —  Reduce  the  fraction  to  its  simplest  form,  and  find  the 
cube  root  of  each  term  separately. 

STOCKS  AND  BONDS. 

Capital  Stock  is  the  money  or  property  employed  by  a  corpora- 
tion in  its  business. 

A  Share  is  one  of  the  equal  divisions  of  capital  stock. 

The  Stockholders  are  the  owners  of  the  capital  stock. 

The  Par  Value  of  stock  is  the  face  value. 

The  Market  Value  of  stock  is  the  sum  for  which  it  may  be  sold; 


DEFINITIONS,    PRINCIPLES,   AND   RULES  XXV 

Stock  is  at  a  premium  when  the  market  value  is  above  the 
par  value  ;  at  a  discount,  when  below  par. 

Bonds  are  interest-bearing  notes  issued  by  a  government  or -a 
corporation. 

A  Dividend  is  a  percentage  apportioned  among  the  stockholders. 
A  Stock  Broker  is  a  person  who  deals  in  stocks. 
Brokerage  is  a  percentage  allowed  a  stock  broker  for  his  services. 
In  Stocks  and  Bonds, 

The  par  value  is  the  base. 

The  rate  per  cent  premium,  or  discount,  is  the  rate. 

The  premium,       ^ 

discount,  or  >  is  the  percentage. 
dividend       J 

f  amount,  or 
Terence. 


The  market  value  is  the  <   ,.^ 
{diffe 


NOTES,  DRAFTS,  AND  CHECKS. 

A  Promissory  Note  is  a  written  promise  to  pay  a  specified  sum 
on  demand,  or  at  a  specified  time. 

The  Face  of  a  note  is  the  sum  named  in  the  note. 

The  Maker  is  the  person  who  signs  it. 

The  Payee  is  the  person  to  whom  the  sum  specified  is  to  be 
paid. 

The  Indorser  is  the  person  who  signs  his  name  on  the  back  of 
the  note,  thus  becoming  liable  for  its  payment  in  case  of  default 
of  the  maker. 

An  Interest-bearing  Note  is  one  payable  with  interest. 

If  the  words  "  with  interest "  are  omitted,  interest  cannot  be 
collected  until  after  maturity. 

A  Demand  Note  is  one  payable  when  demand  of  payment  is 
made. 

A  Time  Note  is  one  payable  at  a  specified  time. 

A  Joint  Note  is  one  signed  by  two  or  more  persons  who  jointly 
promise  to  pay. 


XXVI  SUPPLEMENT 

A  Joint  and  Several  Note  is  one  signed  by  two  or  more  persons 
who  jointly  and  severally  promise  to  pay. 

In  a  joint  note,  each  person  is  liable  for  the  whole  amount, 
but  they  must  all  be  sued  together.  In  the  joint  and  several 
note,  each  is  liable  for  the  whole  amount,  and  may  be  sued 
separately. 

A  Negotiable  Note  is  one  that  may  be  transferred  or  sold.  It 
contains  the  words  "  or  bearer,"  or  "  or  order." 

A  Non-negotiable  Note  is  one  not  payable  to  the  bearer,  nor  to 
the  payee's  order. 

The  Maturity  of  a  note  is  the  day  on  which  it  legally  falls  due. 

A  Draft,  or  Bill  of  Exchange,  is  a  written  order  directing  the 
payment  of  a  specified  sum  of  money. 

The  Face  of  a  draft  is  the  sum  named  in  it. 

The  Drawer  is  the  person  who  signs  the  draft. 

The  Drawee  is  the  person  ordered  to  pay  the  sum  specified. 

The  Payee  is  the  person  to  whom  the  sum  specified  is  to  be 
paid. 

A  Sight  Draft  is  one  payable  when  presented. 

A  Time  Draft  is  one  payable  at  a  specified  time. 

An  Acceptance  of  a  time  draft  is  an  agreement  by  the  drawee 
to  pay  the  draft  at  maturity,  which  he  signifies  by  writing  across 
the  face  of  the  draft  the  word  "  accepted  "  with  the  date  and  his 
name. 

A  Check  is  an  order  on  a  bank  or  banker  to  pay  a  specified 
sum  of  money. 


ANSWERS.  —  Part  III. 


Page  448. 

2.  20  and  80. 

3.  $2000; 

$4000; 
1 12,000. 

4.  18  girls;  36 

boys. 

5.  13  and  65. 

6.  13. 


9. 
10. 


11. 

12. 
13. 

14. 
15. 
16. 
17. 


Page  449. 

.   11. 

i.    $3000; 
$6000; 

$  18,000. 
12  and  60. 
9  marbles  ; 

18  marbles ; 
27  marbles. 
36  years ; 

6  years. 
8. 
1;     4;     12; 

24. 

30;  15;  135. 

9  pounds. 

19  rods. 
85  feet. 


Page  450. 

18.   Son,  $  40  ; 
daughter, 
$80. 


19.  25  days. 

20.  Girl,  $80; 

boy,  $  40. 

21.  Father,  30  da.; 

son,  15  da. 

22.  3    dimes,    6 

nickels,     18 
cents. 

23.  25  yards. 

24.  25    rods;    100 

rods. 

25.  Speller,  15  f.-, 

reader,  45  f. 

26.  60  and  12. 

27.  18  nuts  ;  9 

nuts ;  27  nuts. 

Page  452. 

1.  24. 

2.  24. 

3.  42. 

4.  84. 

5.  24. 

6.  70. 

7.  72. 

8.  40. 

9.  360. 

10.  160. 

11.  18. 

12.  18. 

13.  8. 

14.  16. 

15.  12. 


16.  20. 

17.  900. 

18.  60. 

19.  60. 

20.  32. 

Page  453. 

21.  36. 

22.  222. 

23.  180. 

24.  72. 

25.  320. 

26.  7. 


1.  15  and  75 

2.  28f;  71^-. 

3.  $816. 

4.  $180. 

5.  89. 

6.  100. 

7.  40;  15. 

8.  f|. 

9.  %\. 

Page  454. 

10.  60;  420. 

11.  540;  18. 

12.  9. 


16.  $300. 

17.  64  marbles. 

18.  $2;  $3;  $10. 

19.  $4;  $2. 

20.  3  horses;    12 


Page  455. 

1.  19. 

2.  22. 

3.  47. 

4.  14. 

5.  9. 

6.  10. 

7.  6. 

8.  33. 

Page  456. 

9.  27. 


10. 
11. 


12.  96. 

13.  144. 

14.  18. 

15.  24. 

16.  6. 

17.  32. 

18.  18. 


13.  20  peaches;  5  19.  12. 

plums.  20.  20. 

14.  $200;   $600;  

$700.  2.  15. 

15.  $60;  $140.  3.  9. 


2 

ANSWERS. 

4.  15 marbles;  33 

i    7. 

$  10.32. 

41. 

18,400. 

4. 

Selling  price, 

marbles. 

8. 

$1255.80. 

42. 

40%. 

$831.25. 

9. 

$9.16. 

43. 

133f 

5. 

Selling  price, 

Page  457. 

10. 

$3.68. 

44. 

72. 

$  1051.38. 

5.   25  ft.;  100  ft. 

11. 

$  1.60. 

45. 

68. 

6.   39  acres  ;  47 

12. 

$  1.58. 

46. 

75% 

Page  465. 

acres. 

13. 

$1.62^. 

6. 

3% 

7.   1059  votes; 

14. 

$  326.80. 

Page  462. 

7. 

33^%. 

1377  votes. 

15. 

3  centa. 

47. 

83.i% 

8. 

15% 

8.   62  years. 

16. 

$  13.09. 

48. 

Ux 

9. 

20% 

9.  84;  12. 

17. 

$19.98. 

200 

10. 

<r[  %• 

10.   $108. 

18. 

$1.17. 

49. 

$800. 

11. 

«1% 

11.    17;  28. 

19. 

$3.16. 

50. 

I 

12. 

20% 

12.    $16;  $11. 

20. 

$  7.50. 

51. 

37i% 

13. 

r>%- 

13.  Cows,  $45; 

21. 

$3.21f. 

52. 

99. 

14. 

12^% 

horses, 

22. 

$11.32. 

53. 

$218.75. 

15. 

H>r/o- 

$125. 

23. 

$63.04. 

54. 

$500. 

16. 

Cost,  $375. 

14.   3    dimes;    14 

24. 

$  19.95. 

55. 

1. 

17. 

Cost,  $92.30. 

half  dimes. 

25. 

$550. 

56. 

r>oo% 

18. 

Cost,  $1234.56. 

15.   74  and  26. 

26. 

$44.40. 

57. 

32%. 

19. 

Cost,  $  240. 

16.   21    boys;    33 

27. 

$323.40. 

58. 

$183.20. 

20. 

Cost,  $63.75. 

girls. 

28. 

$4.65. 

59. 

1100. 

21. 

$55. 

60. 

738. 

22. 

25% 

Page  458. 

Page  461. 

61. 

10,800. 

23. 

$800. 

17.    $3600;  $6000; 

30. 

13.T 

62. 

$1547. 

$  8400. 

20  ' 

63. 

$  764.80. 

Page  466. 

18.    44;   11. 

31. 

40%. 

24. 

50% 

19.   5  five-cent 

32. 

X 

Page  463. 

25. 

14?% 

stamps ;     20    two- 

4" 

64. 

$1120. 

26. 

12^/0- 

cent    stamps ;     35 

33. 

168. 

66. 

$170. 

27. 

37^  cents. 

postal  cards. 

34. 

6.r 

66. 

3500  bn. 

28. 

$  40. 

20.   8  horses;    25 

5' 

67. 

$1440. 

29. 

$219. 

cows  ;  55  sheep. 

35. 

110. 

68. 

$592. 

30. 

$200. 

36. 

2x 

31. 

371% 

Page  460. 

3  ■ 

Page  464. 

32. 

2-^% 

1.  $6.34. 

37. 

117. 

1. 

Selling  price. 

33. 

$1646. 

2.   $4.50. 

38. 

X 

$2157.40. 

34. 

$4053. 

3.  $9.60. 

150 

2. 

Selling  price. 

35. 

$113.75+ 

4.  $1.55. 

39. 

500  o/„. 

$  29.40. 

5.  $4,158. 

40. 

.T 

3. 

Selling  price. 

Page  467- 

6.  $415.80. 

800 

$1181.25. 

1. 

3  feet. 

I 


ANSWERS. 

r 

2.    18  feet. 

19. 

331%. 

6. 

$1.16-. 

4.   m- 

3.   I  acre. 

20. 

12%. 

7. 

$  19.55. 

5.    16o/„. 

4.    1430  yards. 

21. 

$  1.92. 

8. 

$  2.94. 

6.   $7.35. 

5.    8  strips. 

22. 

$  24.24. 

9. 

$1,111 

7 

'.  M- 

23. 

$39.20. 

10. 

$67,721 

8 

•.   33|-  times. 

Page  468. 

24. 

$1.57. 

11. 

$49.77+. 

9 

'.    ^ocent. 

6.    3600  sheets. 

12. 

$  235.751. 

7.    250  ft.  1500 

Page  470. 

Page  475. 

sq.  ft. 

25. 

$  18.33. 

Page  472. 

10. 

1.15;  .000625 

8.    250  boards. 

26. 

11^2^% 

13. 

$  18.88. 

.0040625  ; 

9.   $710. 

27. 

$200. 

14. 

$7.80. 

^^0 ;  tAcj-  ; 

10.   6400  cakes. 

28. 

$  893.20. 

15. 

$18.74-. 

sMo- 

11.   76,800  cu.  ft. 

;  29. 

$1. 

16. 

1 1050.801 

11. 

^  %• 

2208  tons. 

30. 

2%. 

17. 

$  1035.731. 

13. 

15  times. 

12.    160  feet. 

31. 

$18. 

18. 

$33.73.-. 

14. 

1 530. 

13.   281,600    sq. 

32. 

$  626.05. 

19. 

$49.04-. 

15. 

.0223125  mile. 

ft.;    240,- 

33. 

$1.57. 

20. 

$  154.871 

16. 

2.432013984. 

000  sq.  ft. 

34. 

$68.18. 

21. 

$884.53+. 

17. 

3|%. 

14.   $27731 

35. 

^m%- 

22. 

$6191.20. 

18. 

$  2.64-. 

15.   96  lots. 

36. 

14!%. 

23. 

$2841.81-. 

19. 

$  3999.24. 

16.    164sq.  yd. 

37. 

mi  %• 

24. 

$2344.50+. 

20. 

162  days. 

38. 

$  27.20. 

25. 

$161.00+. 

21. 

16f%. 

Page  469. 

39. 

15  cents. 

26. 

$835.31-. 

22. 

$5.69+. 

1.    111%. 

40. 

$6300. 

27. 

$886.17-. 

23. 

$753. 

2.    10%. 

41. 

$871.83. 

28. 

$411.65+. 

24. 

$11.28-. 

3.    111%. 

42. 

$309.14. 

25. 

T%-,j%\- 

4.   $26.40. 

43. 

8%- 

Page  473. 

5.   $2.80. 

44. 

7A%- 

29. 

$1550.21-. 

] 

Page  476. 

6.    $1,221. 

45. 

$21. 

30. 

$118.14-. 

1. 

3613  flags; 

7.    450/0. 

46. 

$677.25. 

l^-  yards  re- 

8.   1350%. 

47. 

$7692. 

Page  474. 

maining. 

9.   41%. 

48. 

$2,36. 

18. 

63  cents. 

2. 

200  men. 

10.  A%. 

49. 

n%- 

19. 

$5.70. 

3. 

317  acres  ; 

11.   25%. 

50. 

$21.55. 

20. 

$4.20. 

$  14,325.591 ; 

12.   3310/0. 

21. 

$  1.50. 

$45.19JAV 

13.   16|o/,. 

Page  471. 

22. 

$2.75. 

4. 

$288.75. 

14.   $75. 

1. 

$112.50. 

23. 

$7.20. 

5. 

7,002,079.- 

15.   21f|o/o. 

2. 

90f  cents. 

003129. 

16.  $35. 

3. 

85f  cents. 

1. 

f- 

6. 

482.9638599. 

17.   $71.34|. 

4. 

$19.52-. 

2. 

71  feet. 

7. 

2.0635+. 

18.   $19.20. 

5. 

$  1.65. 

3. 

10mi.249| 

rd.  8. 

$10.38-. 

ANSWERS. 


Page  477. 

12. 

Wt- 

22. 

58,975. 

7. 

106|  yards. 

9.  5or,%; 

13. 

m- 

23. 

,   899,100. 

8. 

86f  bundles. 

i^nr/o; 

14. 

n- 

24, 

,  426,000. 

9. 

810  sq.  yd. 

30jV%. 

25, 

,    16,800. 

10. 

114  rods. 

10.   $.00703+. 

Page  480. 

26. 

1,172,880. 

$.04648- ; 

1. 

^jhh- 

27 

.   4290. 

Page  488. 

$.0055  ; 

2. 

82. 

28, 

.   71,400. 

11. 

300  panes. 

$.00675 ; 

3. 

$1.26J. 

29 

.  67,716. 

12. 

20,736  gal. 

$.00475 ; 

4. 

113. 

30 

.   293,249|. 

13. 

72  cords. 

$.0225 ; 

5. 

531%; 

14. 

13,440  bu. 

$.005359+. 

46f%. 

Page  485. 

15. 

7000  cu.  yd. 

12.   8272.08512. 

6. 

$1.60. 

"  All  other  arti- 

16. 

2250  cu.  yd. 

13.  8.522. 

14.  3.7857+. 

7. 
8. 

192  planks 

12^%; 

1     I 

'$73,327,274; 
122,469  ; 

.  cies, 

$72,1 

1. 

$313.31-. 

15.   30,000  men. 

1U%- 

Increase,     $26,- 

2. 

$136.50. 

16.  i 

9. 

$17.76+. 

976,455. 

3. 

$500. 

17.   15  miles. 

10. 

$6. 

4. 

8%- 

11. 

266|%. 

Page  486. 

6. 

$410.16-. 

Page  478. 

1. 

^^^tmi 

18.    170  bushels. 

Page  484. 

2. 

6l26|Un- 

Page  489. 

19.    1500  letters; 

1. 

165. 

3. 

ll,202|tHi. 

7. 

29.623-  feet. 

750  letters. 

2. 

252. 

4. 

11,6993|0|;}- 

8. 

$30.87i 

20.   $2630.20. 

3. 

4048. 

5. 

786?i|i2|. 

9. 

12}  tons. 

4, 

910. 

a 

49,167[|ff|. 

10 

*? 

1.    1,163,117,- 

5. 

2594|. 

7. 

X\J. 

o. 

683.002129. 

6. 

5646|. 

8. 

963UIUI. 

1. 

$.41+; 

2.   69,092.- 

7. 

1977f 

9. 

™mmi 

(|.37i). 

80236843. 

8. 

6373J. 

10. 

6462i|HH- 

2. 

$.55+; 

3.    176.303- ; 

9. 

5067tV- 

($.46-). 

2.0247-. 

10. 

8852}. 

1. 

630     boards ; 

3. 

$2.89+; 

4.   $66.45-. 

11. 

46,018. 

140  posts. 

($2.75+). 

12. 

79,520. 

2. 

10  feet. 

4. 

$.37-; 

Page  479. 

13. 

25,554|. 

3. 

$1.50. 

($.32). 

6.  $42.96-. 

14. 

106,908. 

5. 

$6.89-; 

6.  X  3,  etc. 

15. 

65,471. 

Page  487. 

($6.72). 

18G.C.D. 

16. 

65,635^. 

4. 

4840  sq.  yd.  ; 

7.   25%. 

17. 

86,855. 

about      70 

Page  490. 

8.  m% 

18. 

27,702^. 

yards. 

6. 

$173.59^ 

9.  m 

19. 

37,411. 

5. 

400  rods. 

($173.70). 

10.    88VW  t'-ents. 

20. 

26.969. 

6. 

18 sq.ft.;  270 

7. 

$36.23+ 

11.   7.625. 

21. 

41,382. 

cu.  ft. 

(1 36.26+). 

ANSWERS. 


8. 


10. 


11. 


12. 


13. 


14. 


15. 


16. 


17. 


$791.42+; 
($791.89-). 
$  176.741; 
($  176.85). 
$  985.411 ; 
(.f  986). 
13.10; 
(!?2.80). 
$2.60; 
($  2.48). 
$5.64; 
($5.46). 
$  1.74-  ; 

($  1.69+). 
$  5.38- ; 
($5.29+). 

$2.52; 

($  2.40). 

$5.58; 

($5.40). 


18 


Page  491. 

Proceeds, 
$87.34-; 
($87.38+). 
19.  Proceeds, 
$122.66+; 
($122.75-). 
Proceeds, 
$  502.05-  ; 
($  502.34+). 
Proceeds, 
$71.65-; 
($  71.68+). 
Proceeds, 
$  232.99- ; 
($233.10+). 
Proceeds, 
$95.58+; 
($95.63-). 


20. 


21. 


22. 


23 


24. 


25. 


Proceeds, 
$162.65+; 
($162.76+). 
Proceeds, 

$81.91+; 

($81.95+). 


Page  495. 

.01020201 
10.01. 


$  24.28. 
$10.15+. 
1 2.52  f. 
$  367.42- 
$  10.99+. 
$2.13. 
$11.23+. 
$12.78+. 


Page  496. 

2. 
3. 
4. 


Page  492. 

674.7+  yards. 
$613.98+. 

Page  493. 


1. 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 


371|. 
5301. 

129rV- 

1272f. 

517^ 

289^. 

233||. 

260ff. 

73413. 

10781. 

85^. 

89f. 

264if. 

361  If. 

337H. 

674H. 

401,V 

421H- 
434i. 


5. 

6. 

7. 
8. 

9. 
10. 
11. 


12. 


$  4.90. 
$115.54-. 
$5.37.1; 
$224  07-. 
$728; 
$950f. 
$  25,000. 
20%. 
Tlie  lirst ; 
5  %  more. 
1|  acres. 
.0006216. 
$1200; 

$  4608  ; 

25%. 


Page  497. 

13.  $55.91-. 

14.  43V; 

6 If  days ; 
$85,3331. 

15.  $173,668; 
$201.880618. 

17.  $1453.76. 

18.  4.23+  times  ; 
64.68+    in- 
habitants ; 

6.07-  inhabi- 
tants ; 
$  15,124,032. 

19.  300  sheep. 

20.  $8572.20. 


Page  498. 

21.    257sq.  yd. 
6i%; 
$287.50. 
52  weeks. 
33,750  qt.; 
2250  qt. 

average. 
$153.75. 
$  33,519.20. 
$85. 
70  feet. 
$  187.36. 
9  J^  0/  • 

«^%; 
100%. 


25. 
26. 

27. 
28. 
29. 
30. 


Page  499. 


$  46.80. 
$  18.40. 
17  lb.  11  oz. 
5  pwt.  19 


5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 


Page  500. 

I.    23,220    gr. 

gold; 
2322  gr. 

silver  ; 
258  gr. 

copper. 
24  spoons. 
$  2800. 

iff;i09fX 

$  105. 
27+  cents. 
23t\  gal. 
704  sq.  ft. 
If  cu.  ft. 
1317U  lb. 


6 


ANSWERS. 

14.   13,405^ 

2. 

$175.94-; 

17.  $45;    $135; 

19. 

$870.48. 

gal-; 

($175.94-). 

$270; 

20. 

A,  $50;  B, 

1440  bu. 

3. 

$  350.55+ ; 

$450. 

$90;      C, 

15.   180  o/o. 

($  350.55+). 

18.  $237;  $189; 

$110. 

16.   ^2}%. 

$114. 

21. 

$380. 

17.  $309.42. 

19.   7  days. 

22. 

$  158.40. 

Page  504. 

20.   240  eggs. 

Page  501. 

4. 

$846.26-; 

21.   5  months. 

18.   300  pounds. 

($845.77+). 

22.    24  weeks. 

Page  509. 

19.  A,  43^%; 

5. 

$724.85+; 

1. 

$554.23. 

B.  36o/o; 

($724.69-). 

Page  506. 

2. 

$171.20. 

0,20^0/^ 

23.  $37,033,131. 

3. 

$1782.67i 

20.   672  yards. 

1. 

38  rd.  3  yd. 

24.    10  inches. 

4. 

$158.40. 

21.  $315. 

11  in. 

25.   $360. 

5. 

$  28.50. 

o 

113  rd.  3  yd. 

1  ft.  6  in. 

a 

$60.56^. 
$50.85. 

Page  502. 

/i. 

1.   125,422,928,- 

o. 

7. 

1;   101.5901+. 

3. 

175  rd.  1yd. 

368.01. 

8. 

$13. 

2.  n\. 

1  ft.  6  in. 

6.    12  T.  6  cwt. 

9. 

$7.45. 

3.   72. 

4. 

£16     13  s. 

2  qr.  13  lb. 

4.  390. 

4d. 

Page  510. 

5.   3  yards. 

5. 

21,090d. 

Page  507. 

10. 

$27.84-. 

6.   $4. 

6. 

£  26  5  s. 

7.   300. 

11. 

40  and  10%; 

7.   20|o/,. 

7. 

$27.37+. 

8.    11  f  days. 

$2  diflfer- 

8.  $81.25. 

8. 

£108    4  s. 

9.   3416^1  lb. 

ence. 

6d. 

£45  188.  8d. 

10.   £3068  158. 
10  d. 

12 

$60;     40  % 
discount; 

1.   llf  rolls. 

9. 

X/6. 

2.   1.5548—. 

10. 

773t?^  oz. 

60%   net. 

3.  5333^  bu. 

11. 

$175. 

4.  302. 

13. 

52%. 

4.  $3.02^f. 

6.   ll^V 

14. 

$100. 

Page  505. 

7.   9|. 

15. 

72% 

Page  503. 

12. 

$123. 

9.    1250. 

16. 

72%. 

5.   $60.62^. 

13. 

$72; 

6.   70  cents. 

$119.25; 

Page  508. 

1. 

$81. 

7.   IW^cta. 

$92.25. 

10.   lOf^f. 

2. 

75%.. 

8.   17  spoons. 

14. 

$450; 

11.  37^  pieces. 

3. 

1^%. 

9.  48/jbu. 

$750; 

12.  43,200  min. 

4. 

627.5. 

$600. 

13.   24f  inches. 

5. 

.0075 ;  ^. 

1.    $49.01-; 

15. 

24  days. 

14.   362.16  mi. 

6. 

77r/o- 

($48.99-); 

16. 

2|     hours ; 

15.   $6.72-. 

7. 

$  2.63f 

$48.88-; 

48     miles 

17.  $4761.90+. 

8. 

$24. 

($48.88-). 

from  A. 

18.   $8.28J. 

9. 

600. 

ANSWERS.  7 

Page  511.  25.    $4.16f.  ^^    1969  a;.  5.  $1.43-. 

10.  $3575.  26.    80  hours.  *     2000  '  6.  $36. 

11.  6|%.  27.    250cu.  ft.  (]^1^\  7.  1  yr.  6  mo. 

12.  $65.  28.    $7.12^.  V  200  )'  8.  $42.17+. 

13.  $101.50.  29.    $18.  9.  $5000. 

14.  $375.  30.    $77.95,  Page  517.  10.  6%. 

15.  $10.96-.  ($79).  1.    121icu.  ft.  11.  1  yr.  9  m 

16.  $23.14-.  2.    $162.  2  da. 

J     9x  12.  $144. 

Page  512.  *    80 '  13.  $83.26i. 

1.  $  19.  2.  $  1600.  Page  518.  14.  3  %. 

2.  35.  3.    255  a;.  3.    $126.  15.  2  yr.  1  mo. 

3.  2624f  yards.  4.    3%.  4.    $85.50.  7  da. 

4.  $27.20  gain.  5.    $2000.  5.    16.8  tons.  16.  $80. 

5.  $20.  6.   2  pieces,  17.  $2181.99-. 

6.  1767.5.  12x12;  18.  $72. 

7.  $5.49-.  Page  515.  2  pieces, 

8.  35|%.  6.    5  years.  12x14;  Page  521. 

9.  $1431.27.  7.    150a;.  2  pieces,  19.  5  mo.  23  da. 


8.  6%.  14x14.  20.    6%. 

9  1875  4-^2^  '^'    44iflb.  21.    $16.92. 

Page  513.  *  32  8.    5280  lb.  22.    $  402.22. 

10.  $30.  10.  682.90+  9.    Outside  di-  23.    37  days. 

11.  4/y.  6829  a;  mensions,  24.    5%. 

12.  .00007865.  3200  '  14x14x14;  25.    11  mo.  29 da. 

13.  108.86-  bu.      jj  21^ .  2744  cii.  m. 

14.  2bu.lpk.4qt.        *  2000'  wood  and  Page  522. 

15.  $7425.  /_x_\  marble;  1.    570,073,438,- 

16.  25%.  \100J'  1728  cu.  in.  098.53. 

17.  $49.31+.  j2  11a;.  fx\  marble 


^^(f) 


18.  $594.50;               '     30  '  V3/  1016  cu.  in.  Page  523. 

($594.80).  j3    x+3_  f  x\  wood.  6.    .37875. 

19.  25%;32j%;        ■     20  '  Uoj'  10.    8  times ;  7.   $70.20. 

42f%.  ^4    ^.(ll]  iton;  8.    $281.25. 

20.  $1.56|.                  *     60  'V  3/  6|tons.  9.   $15,000. 

21.  Loss,  $177.  ^g    5997-x .  10.   $67.71-. 

22.  $3.60.                   *       10      '  Page  520. 

23.  .Olf.                         (qoO-^\  ^'    •'''^'-  Page  524. 

V            10/  2.    2  years.  2.    i?  41.99+. 

Page  514.  ,g    477x  3.    $96.  3.   $951.13+. 

24.    15,203.                '     400  ■  4.   S%.  4.    $112.74-. 


8  ANSWERS. 

5.  $119.43-.  34.    I6.55+.  7.  4269.22+  11.    |853.27+; 

6.  $13.91-.  35.    $278.16-.                    francs.  ($853.71-). 

36.   $1.23-.  8.  $1563.55-.  12.   $56.62^; 

Page  525.  37.   $196.64+.  9.  $1563.55-.  ($57.50). 

7.  $147.19-.  38.   $1.10.  10.  $1547.37. 

8.  $8.10-.  39.  $389.60.  ^^  18^  Page  536. 

9.  $52.33-.  40.   $6.11+.  '25'  3.   2^^. 
10.   $1005.50.  41.   $4.09+.  4.   375. 

12.  $967.78.  42.  $56.32-.  5.   $288. 

13.  $21.79-.  43.   $16.72-.  Page  531.  g.   $922.20. 

14.  $16.53-.  44.    $95.43-.  12.  400-4a;.  7.   $573.47^ 

15.  $1274.21-.  45.   $15.71+.  13.  $500.  8.   $1700. 

46.  $594.30;  14.  10%.  9.   $1400; 

Page  526.  ($594.60).  16.  Same.  $1200. 

16.  ($800.50-).  47.   $10.73+;  10.    $1337. 

17.  $24.74-;  ($10.38+).  1.  688,965,549,- 

($23.94).  48.    $793.73+;                  176.65.  Page  539. 

18.  $761.06-;  ($794.13+).  1.    $165.37^. 
($761.45).  Page  532.  2.   $1453.42-. 

19.  $43.99-.  Page  528.  6.  .0019.  3.  $50,625.50. 

20.  $786.39-.  49.    45  cents ;  7.  32  cents. 

21.  $65.40.  (nothing).  8.  $46  gain.  Page  540. 

22.  $625.03+.  50.    $968.83+;  9.  60  cents.  4.  $893,615,929. 

23.  $98.49+.  ($970).  10.  $16.98.  7.   284,106,409,- 

24.  $993.27+.  352.02. 

25.  $61.68-.  1.   470,952.  Page  535.  8.    45.29-%. 

26.  $252.37+;  1.  $106.33+;  9.    $223,852.8835. 
($252.52-).  Page  529.  ($106.38)-.  10.    107§  cents. 

27.  $13.09-;  5.    8614.20375.  2.  7%; 

($12,871).  6.    2f  {ri^%).  Page  541. 

28.  $486.10-;  7.    $110.85+  3.  24  days.  11.    $20,000. 
($486.43-).  gain.  4.  $1200.  12.    $728.17^. 

29.  $2.33+;  8.    $16.25.  5.  $4.68;  15.    Profits, 
($1.94+).  9.   45T.  5cwt.  ($3.90).  $4414.10. 

30.  $989.67-.  2qr.  6.  5%.  16.    |;  |;  t^;  f 
($990).  10.   $43.33^.  7.  72  days. 

31.  $1938.43-.  8.  $1200.  Page  542, 

32.  $8.06+.  Page  530.  ($1260).  17.   $4752. 

4.   $196.17.  9.    $304.26-.  18.    37'.%;  25%; 

Page  527.  5.    $600.01+.  ($304.05+).  12^%;  25%. 

33.  $1473.52-.  6.   $1506.12.  10.  Apr.  8,  1894.  19.    |81.90. 


ANSWERS. 

9 

20.   $299.88-. 

38. 

24,130f. 

8. 

$3. 

9. 

AiiV 

39. 

651, 329  j\. 

9. 

£73  3s. 

10. 

Largest,    f ; 

1.    1,287,400. 

40. 

1,932,560. 

10. 

6f  days. 

smallest. 

2.    3,370,185. 

41. 

$277,133.11. 

foff. 

3.    598,969. 

42. 

$60,887.10. 

2. 

$58.93-. 

4.    2,883,736. 

43. 

$48,554.08. 

3. 

6%- 

Page  550. 

5.    816,669. 

4. 

$3400; 

11. 

2^. 

6.   5,127,460. 

Page  544. 

$3570. 

12. 

2  da.  15  hr. 

7.    2,455,038. 

44. 

£7  5s.  9d. 

50min.  35 

8.   42,327,198. 

45. 

11  yd.  1  ft. 

Page  547. 

sec. 

9.   2,513,420. 

11  in. 

5. 

$1.30-. 

14. 

iViV 

10.    22,944,747. 

46. 

12  bu.  2  pk. 

6. 

$  10,000. 

15. 

ItV 

5qt. 

8. 

$264.25. 

16. 

f^- 

Page  543. 

47. 

11.76+. 

9. 

$  135.40-. 

17. 

180. 

11.    857,712. 

48. 

13.72+. 

18. 

3f. 

12.    6,482,112. 

49. 

14.41+. 

Page  548. 

19. 

t¥^. 

13.    1,230,828. 

50. 

13.34+. 

1. 

$107.65+. 

20. 

36f. 

14.    921,776. 

51. 

19.05-. 

2. 

6  yr.  6  mo. 

21. 

ItV 

15.    3,460,704. 

52. 

22.30-. 

3. 

n%- 

22. 

3hr.22min. 

16.    5,888,304. 

4. 

$750. 

23. 

85i  rods. 

17.    1,460,025. 

1. 

79.98  %. 

5. 

$882. 

24. 

iM- 

18.    10,563,960. 

2. 

5.02%. 

6. 

$97. 

25. 

15f. 

19.    3,911,322. 

3. 

4.28%. 

7. 

$6;  $144. 

20.    2,982,840. 

4. 

3.83  %. 

8. 

33r/o- 

1. 

4|  feet. 

21.    714,186. 

5. 

3.81%. 

9. 

$  8960. 

2. 

120  yards. 

22.    3,277,719. 

6. 

2.44%. 

10. 

$360. 

23.    456,375. 

7. 

.54%. 

Page  551. 

24.    174,600. 

8. 

•10%. 

Page  549. 

3. 

8750  sq.  yd. 

25.   362,250. 

Total,  $872,- 

11. 

$50. 

4. 

3000+ 30  ,t; 

26.    104,787i 

270,283. 

12. 

20. 

80  yards. 

27.    128,550. 

13. 

$61.87|. 

5. 

100  a; ;  40 

28.    625,975. 

Page  545. 

14. 

$1150. 

yards. 

29.    213,966f. 

1. 

237.49%. 

15. 

^%- 

6. 

60  a; +1200; 

30.   413,866f. 

2. 

234.60  %. 

80yd.  and 

31.    86501 

3. 

563.36  %. 

1. 

4ff 

*120  yd. 

32.    10,757f. 

3. 

tV- 

7. 

10,000  sq. 

33.    21,873^. 

Page  546. 

4. 

1611. 

yd. 

34.   24,292i. 

4. 

0.04%. 

5. 

m- 

8. 

1920  flag- 

35.   485,072. 

5. 

25%. 

6. 

m 

stones. 

36.   167,276. 

6. 

20%. 

7. 

73;  240. 

9. 

$9.26-. 

37.    24,418^. 

7. 

$24;  $30. 

8. 

m- 

10. 

46 1  yards. 

10 


ANSWERS. 


11.  90^1  lb. 

12.  35/^  sq.  ft. 

Page  552. 

13.  10  ft.  8  in. 

14.  950  bushels. 

15.  810  gallons. 

16.  $416. 

17.  10  miles. 

18.  615  cords. 

19.  Bj-mib. 

20.  27/^  lb. 

21.  70G|sq.  j^d. 

22.  722f8q.  yd. 

Page  553. 


1.  24H  yards. 

2.  28  cents. 

Page  554. 

3.  84  days. 

4.  9  days. 

5.  $49. 

6.  180  bushels. 

Page  555. 

8.  (i()  husliels. 

9.  720  pounds. 

10.  4G00  sheets. 

11.  $312. 

12.  8  da.  4  hr. 

13.  !?  2^.80. 

14.  9()  rods. 

15.  50  days. 

16.  $45. 

17.  120  men. 

Page  556. 

18.  360  men. 


19. 
20. 

2. 
3. 
4. 
5. 
6. 
7. 


10  hours. 
$1.20. 


329. 

10  bushels. 

12  feet. 
90  cents. 


Page  557. 

8.    1680. 
85,800. 
Loss, 
$93.75. 

$2.35+. 
99.99. 
14  feet. 
$69.05-. 
1050  acres. 
9792f  lb. 
$  275.69-. 


9. 
10. 

11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 


Page  558. 

20.  $873; 
$184; 
$143. 
Estate, 

$  45,000 ; 
son's   share, 

$  10,000. 
$10.86+. 
6^  days. 


21. 
22. 


23. 
24. 


$78;  $104. 
$36;    $17 

$48. 
$25;  $20. 

$16.20; 

$  13.80. 


5.  45  lb.  gold  ; 
4^  lb.  silver; 
I  lb.  copper. 

Page  559. 

6.  ISaltpeter, 

54  lb.  ; 
sulphur, 

7ilb; 
charcoal, 
10^  lb. 
$  20.25 ; 
$  18.63  ; 
$15.12. 
864  bales 
540  bales  ; 
396  bales. 
$12;  $20; 
$60;  $72. 


7. 


8. 


20.  5934.47- 

meters. 

21.  6  quarts. 

22.  .8125  pound 

23.  .25  rod. 

24.  100  links. 

25.  .1  acre. 

Page  561. 

£13138. 

8|-d. 
368.90 

marks ; 
$30.73-. 

Page  562. 

1.  $40. 

2.  8o/o. 

3.  $150;  150% 


2.  326.9843. 

3.  6408. 
5.   4.2633; 

1.405712. 


Page  560. 
6.  16.5393. 
9.    28.165; 

305.36721612. 
10.  £.3375. 
12.    Sum, 

8.08690625. 

14.  .019104141. 

15.  .05;  .03965. 

16.  1093.3524 

17.  2.8. 

18.  .215625  m. 

19.  12  da.  20  hr. 

31  min.  12 
sec. 


Page  563. 

4.    $30. 

^%- 
$435. 
Same. 


5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 


f)  0/  .      •)  1  0/ 
'-'   /o  •     -2    /o- 

$2267.26+. 
$2.85. 


Page  564. 

17.  .f  409.37 i. 

18.  $24,000*"; 


19. 
20. 
21. 

22. 


u%. 

$  18,000. 
101  rd.  2  yd. 

1  ft.  6  in. 
35    shares ; 

$87.60. 


ANSWERS. 

11 

23.    37  rd.  4  yd. 

Page  570. 

26. 

$  57,600. 

2. 

$1874.771. 

11  in. 

1. 

$45. 

27. 

$2,371 

3. 

$2461.76+. 

24.    $106.12. 

2. 

10|  acres. 

28. 

$387.36. 

4. 

$  1000.50+. 

3. 

$452,321 

5. 

$  1632. 

4. 

$  4.50. 

Page  573. 

Page  565. 

5. 

$1.50; 

29. 

$1000; 

25.    229     rd.     4 

$5.40; 

100%. 

Page  577. 

yd.    2    ft. 

20%; 

30. 

$  5487. 

6. 

$  1845.90+. 

6  in. 

6. 

$  6000. 

7. 

$  946.04-. 

27.  $428.40-; 

7. 

$3333^. 

2. 

688,450. 

8. 

$  632.65-. 

$53.40-. 

3. 

3f f  ;  18|f . 

9. 

$985.22+. 

28.   $119.10+. 

4. 

.140625; 

10. 

$326.34. 

29.    $149.14+. 

Page  571. 

88.088. 

8. 

$212.51-; 

5. 

5.820068 ; 

1. 

9  hr.  45  min. 

$439.79-. 

1000. 

2. 

14|  minutes 

Page  566. 

9. 

$  5250  ; 

past  5  P.M. 

30.    51  cents. 

18%. 

Page  574. 

3. 

4  ft.  6  in. 

10 

$  402.50  ; 
35%. 

B 

400  yards. 
$2f. 

1.    15%. 

xv. 

7. 

Page  578. 

2.    $400. 

11. 

$  3330 ; 

8. 

$  10,500. 

4. 

748  plants ; 

3.    $102.50. 

$4.25. 

9. 

7400  inhab- 

754 plants. 

4.    $2.70. 

12. 

$9.30. 

itants. 

5. 

20  da.  6  hr. 

5.   $40. 

13. 

$201.60; 

10. 

$  10,500. 

40  min. 

6.   $1.60. 

$125.93-; 

6. 

5544     revo- 

7.   1  hour. 

$  722.40+. 

Page  575. 

lutions  ; 

8.    $20. 

14. 

2|  pounds. 

1. 

(a)  1575.- 

1320     revo- 

9.   $12. 

15. 

98t¥3  lb. 

355671 ; 

lutions. 

16. 

$71.25. 

(6)  .028376- 

7. 

28  mi. 

Page  567. 

604. 

130if  rd. 

10.    $20  loss. 

Page  572. 

2. 

49,999.- 

8. 

1388|  mi. 

17 

$85. 

$  594.20 ; 

74000 

Q 

15708  f^n^ 

4.    1.. 

X  1  . 

18. 

3. 

(a)  73  ; 

J7. 

10. 

694f  miles. 

5.    962  feet. 

($594.50). 

(b)  2016. 

11. 

3211i|  mi. 

6.    $6.50  gain. 

19. 

20% 

4. 

$72. 

7.    $3.54. 

20. 

$11,356,011 

5. 

$53.95+. 

Page  579. 

21. 

700. 

6. 

$160;   $140; 

12. 

1104f|  mi. 

Page  568. 

22. 

$38.59+. 

$240. 

8.   $509.25. 

23. 

12. 

7. 

$262.50. 

1. 

$828.45. 

9.    19  T.  62  lb. 

24. 

$  1006.13^. 

($828.87). 

8  oz. 

25. 

63  sq.  ft. ; 

Page  576. 

2. 

$397.30; 

10.   24  cents. 

lO^f  rods. 

1. 

$3481.07+. 

($397.50). 

12 


ANSWERS. 


3.  f  554.40; 

($554.68). 

Page  580. 

4.  $625.33+. 
g     3959  a;. 

4000  ' 
/3961a;\ 
V4000  j' 


6. 


7. 


Il979-2a;. 

10 
/11985-2a;^ 
V        10        > 

7956 +  8  a; 


(1592. 1). 

8.  $1200; 
($1199.39+). 

9.  30  days; 
(33  days). 

10.    $1.50  discount 
per  $1000. 
($2  discount 
per  $  1000). 

Page  581. 

1.  56°. 

2.  2  hr.  29  min. 

12  sec. 

3.  5  hr.  50  min. 

20  sec. ; 

7  A.M. 

4.  37°  30^. 

5.  1  hr.  14  min. 

52  sec. 

Page  582. 

6.  33°  30^  east 

longitude. 

7.  39^gf^    miles 

per  hour. 


8. 

9. 
10. 
11. 
12. 

13. 
14. 


1800  lb. 
$  1000.80. 
8  yd.  6  in. 
15  ft.  7  in. 
257bu.2pk. 

2  qt. 
20  spoons. 
16°  40'. 


Page  583. 

1.  $67.46; 
$  168.65 ; 
$236.11. 

2.  1  lb.  3  oz. 

3.  .93. 

4.  1  yr.  4  mo. 

26  da. 
nearly. 

5.  16  days. 
7.   .V^. 


20.  /j;  ^; 

21.  1.299609375. 
23.    109|ffeet. 


Page  585. 

24.  A,  4340 

votes ; 
B,  5551  votes. 

25.  $133.32-. 


13. 
14. 

15. 

16. 


17. 

18. 
19. 


C,  119|. 

3  bu.  3 

3qt. 

$7.87i. 


pk. 


1.  $880.86+. 

2.  $1229.01-. 

Page  586. 


Page  584. 

8.  $630.76-. 

9.  46f  inches. 

10.  .0005207. 

11.  yW  acre. 

12.  30  min.   50 

sec. 
$225.67-. 
Dec.  (10)  13, 

1888. 
Dec.  29, 

1892. 
A,  1071 ; 


Page  587. 

10.    $908.S7-. 

Page  589. 

1.  $224.46+. 

2.  $261.19-. 

Page  590. 

4.  $772.37-. 

5.  $899.91+. 

3.  12. 

4.  2H. 


Page  591. 

$150; 

$7500. 

15,544,041.4.= 
francs. 


1.  $3640. 

2.  $4;  6|o/,. 

Page  592. 

6.  $  2  loss. 

7.  $701.53-. 

8.  $13.65. 

10.   $200.02^. 


$  193.70+. 
$446.33-. 
4188.48+ 

marks. 
8490.44- 

francs. 
£  307  7  s.  6^ 

d.  nearly. 
$2050.72+. 
$4138.97+. 


1.  $137.61+. 

Page  593. 

2.  $12.39-. 


1. 
2. 
3. 

4. 
5. 
6. 
7. 
8. 
9. 
10. 


$8500. 
$282.25. 
23  min.  53^ 

sec. 
$  17.82. 

1 1  feet. 

12  rods. 
$56. 

l\  cent«i. 
$300. 
3600  yards. 


Page  594. 

11.   $497; 
($497.25). 
5%.  IS.f'oCt 
$700. 
$375.10-. 


12. 
14. 
15. 


ANSWERS. 


13 


11,181,021.50. 
$1,037,124.44. 
1214,854.74. 
£  14  7  s.  2  d. 
25  bu.    1   pk. 

4qt. 
23    rd.   2  yd. 

1  ft.  6  in. 


Page  595. 

7.  $249,981.53. 

8.  $318,808.78. 

9.  $202,722.44. 


Page  597. 

$11. 
$85.25. 
$95.48-. 
24  cents. 
$  18,228. 
150  sq.  ft. 
$  39,700. 

138  feet. 


Page  598. 

10.  $48.43|. 

11.  44%. 

12.  $33.12^. 

13.  114sq.  yd. 

14.  $594. 

15.  $2090.25. 

16.  if. 

17.  26|o/,. 

18.  $363. 

19.  64001b.; 

$11.40. 

20.  $870.62-; 

($  870.14-). 

21.  $4.31^. 


22.  45315  ^. 

23.  $150;  $225. 

Page  599. 

24.  $33.60; 
15 A  cords. 


'T^ 


25. 
26. 

27. 


"2    >     ^10* 

10%. 

$  653.48-  ; 

($654.36). 

28.  $2688|. 

29.  $3682.19-; 
($3684.371). 

30.  $4940.28." 


Page  600. 
3.    135.62. 
4      19  3  0 

5.  13  feet. 

6.  16T.  19cwt. 

3   qr.  101 
lb. 

7.  $1235.21-. 

8.  881. 

9.  25  cents. 
10.    $16.48-. 


Page  601. 

1.  1^  yards. 

2.  18.72  feet. 

3.  $78.75. 

4.  8  ft.  4  in. 

5.  $14,910.75. 

6.  108i|  sq.ft. 

7.  44.17875  sq. 

ft. 

8.  2970  bu. 

9.  252  gallons. 
10.    146  sq.  yd. 


11.  3hr.  12min. 
4hr.  6f  min. 
1  hr.  48  min. 

Page  602. 

13.  40^  yards; 
$44,021. 

14.  11   ft.;    108 

sq.  ft. 

15.  984  sq.   ft.; 

71ft. 

16.  $11,200. 

17.  $56. 

Page  603. 

1.  .00007722. 

2.  1.485135. 

3.  .450522. 

4.  7.70904. 

5.  .0712111. 

6.  .0048393. 

7.  .63672. 

8.  .0374715. 

9.  .8220672. 
10.    .00004768. 


1. 

.68515625. 

2. 

589.84. 

3. 

153.6. 

4. 

3265. 

5. 

50. 

6. 

.064. 

7. 

.00002375. 

8. 

.0115. 

9. 

79,000. 

10. 

219.32. 

3.  4.120275. 

Page  604. 

4.  li^^. 


5. 


9. 
10. 


7  fur.  16  rd. 

3  yd.  1  ft. 

9.888  in. 
$4.13. 
$  1900. 
3T.  7cwt,  2 

qr.  17  lb. 

4^  oz. 
$  15,000. 
$6.98-. 


Page  605. 

2.    1,345,595. 


Page  606. 


4. 
5. 
6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 


80  miles. 

.49184. 

$9.27; 

$12.36. 
468  bricks. 
$46.08. 
55  feet. 
$5700. 
$1700. 
$  77.14. 
$  44.48. 
$215.36+. 
$  747.44-. 


Page  609. 

1.  14. 

2.  16. 

3.  18. 

4.  24. 

5.  26. 

6.  36. 

7.  35. 

8.  42. 

9.  44. 
10.  51. 


14 


1 

ANSWERS. 

11.   55. 

2. 

1 4330.80+. 

21.    $4.10; 

7. 

12:1:20  p.m. 

12.   54. 

3. 

4  lb.  tea,  27 

(14). 

8. 

143°  3' West. 

13.   61. 

lb.  coffee. 

22.   $2750. 

9. 

82°  40' West. 

14.   63. 

4. 

|458.16|. 

23.    90  da. 

10. 

11:33:12 

15.    72 

6. 

$8.25. 

24.    18  da. 

A.M. 

16.  75. 

25.    $475. 

11. 

3:37:20  p.m. 

17.    83. 

Page  614. 

($474.76-). 

12. 

88°  4'  East. 

18.    84. 

6. 

Dec.  (12)  15; 

13. 

32°  34'  East. 

19.    91. 

(27)  30  da.; 

Page  616. 

14. 

2:45:42  a.m. 

20.   95. 

$1275.88+; 
($1276.00+). 

1.  567. 

2.  915. 

15. 

23°  West. 

1.    7568. 

7. 

$77.61-. 

3.    144,200. 

3. 

M- 

2.   5107. 

8. 

$332.50; 

4.    25,758. 

6. 

if. 

3.    6008. 

$525. 

5.    2,114,000. 

7. 

if. 

4.    6285^^111. 

9. 

18  men. 

6.    86,526. 

8. 

ff. 

5.    6098. 

10. 

256  barrels. 

7.    106,000. 

9. 

n- 

6.    3007. 

8.    374,625. 

10. 

If- 

7.   98,640. 

1. 

$2167.09-. 

9.    19,800. 

11. 

il 

8.    75,064. 

2. 

$543.16-. 

10.    7312. 

12. 

M- 

9.   70,921. 

3. 

$1335.23+. 

11.   466,000. 

13. 

3h 

10.    78,905. 

4. 

$911.66-. 

12.   50,133. 

14. 

H- 

13.    98,500. 

15. 

If. 

Page  610. 

] 

Page  615. 

14.    4180. 

16. 

2^. 

1.   $183.27-. 

6. 

$  784.54-. 

15.   423,000. 

17. 

m- 

2.    £150  88.  4 d. 

6. 

$  724.03+. 

16.   40,096. 

18. 

33^. 

7. 

$1586.88-. 

17.   419,904. 

19. 

12^ 

Page  612. 

8. 

$1008.10+. 

18.    310,050. 

20. 

16^. 

1.  iff- 

9. 

$2607.48-. 

19.    230^7. 

21. 

lOJ. 

2.   I 

10. 

$718.00-. 

20.    227i 

3.    ,«tV 

11. 

4^%- 

Page  619. 

4.  m- 

12. 

$600. 

1.    12:6  p.m. 

1. 

666,862,394,- 

5.  Mi 

13. 

4  yr.  6  mo. 

588.21. 

6.    f 

9  da. 

Page  617 

5. 

m- 

14. 

2yr.8mo.lda.    2.    9:36  a.m. 

6. 

$1144.73-. 

Page  613. 

15. 

$  600. 

3.    142°  55' 30'^  7. 

$5166.69+; 

7.  mi 

16. 

4%. 

West. 

$283.31-. 

8.    Equal. 

17. 

63  days. 

4.  52°  36' East. 

9.    ^^. 

18. 

$400. 

5.    5:16  a.m. 

Page  622. 

10.    ^. 

19. 

$  295.35. 

Table. 

($295.50). 

Page  618. 

1. 

63.30  %. 

1.   1389.61-. 

20. 

7%. 

6.    8:24  a.m. 

2. 

13.14  o/„. 

3. 

8.58%. 

7. 

M.1M  O  VV 

Page  629. 

18. 

50% 

4. 

3.83%. 

8. 

h 

6. 

60  cents. 

19. 

4% 

5. 

3.41% 

9. 

If- 

7. 

20% 

20. 

$372. 

6. 

2.09%. 

10. 

9 
32"- 

8. 

$150. 

21. 

Iff  days. 

7. 

1.11%. 

11. 

1. 

9. 

$120. 

8. 

.72%. 

12. 

Iff- 

10. 

20%. 

1. 

nu- 

9. 

3.820/,. 

13. 

6. 

2. 

163 

3T5- 

All  others, 

1. 

254. 

126,632,801. 

Page  626. 

2. 

27.1. 

Page  632. 

2. 

$  625.92. 

3. 

1.37. 

3. 

22T-k. 

Page  623. 

3. 

3968  revolu- 

4. 

26.8. 

4. 

^;  4  s.  6d. 

1. 

63,805,573.34. 

tions. 

5. 

3.76. 

5. 

13t\V 

2. 

$  869,109.89. 

4. 

27  kilo- 

6. 

.838. 

6. 

8 f 0  acre ; 

3. 

524,658,551,- 

meters. 

7. 

16.27. 

218  lb.  12  oz. 

760. 

8. 

.4876. 

7. 

A;H- 

4. 

986ff|f. 

Page  627. 

9. 

.306. 

8. 

1 

8- 

7. 

(a)  $1104.02+. 

5. 

770  meters  ; 

10. 

.069. 

9. 

42ff. 

(6)  $1101.56+; 

42  kilos. 

10. 

1. 

$1116.44+. 

6. 

$  504. 

Page  630. 

11. 

10. 

7. 

8|  days. 

1. 

$6.21. 

12. 

£f;  j\  da 

Page  624. 

8. 

$82.50. 

2. 

4hr.  24  niin. 

13. 

8. 

8. 

$9862.33+. 

9. 

14  yards. 

3. 

$  22.50. 

14. 

13  s.  3i}d. 

9. 

f 

10. 

$  2.45. 

4. 

150  barrels. 

15. 

t¥iV- 

10. 

$  72.06. 

5. 

$153.60; 

16. 

3,3,. 



Page  628. 

$192; 

17. 

370. 

1. 

80. 

21. 

40  and  10%. 

$  230.40. 

18. 

$520. 

2. 

4. 

22. 

50  and  10%. 

6. 

1740  tiles. 

19. 

$918.75. 

3. 

17i 

23. 

30  and  10%. 

7. 

$600;000. 

20. 

£1060  18  s 

4. 

28. 

24. 

30  and  5  %. 

8. 

4i%. 

9d. 

5. 

5. 

25. 

30  and  15%. 

9. 

5.0004. 

6. 

7. 

26. 

50  and  10%. 

10. 

2880  gal. 

Page  633. 

7. 

36  cents. 

27. 

20  and  50%. 

11. 

.013972. 

1. 

49|fi  cents 

8. 

70  cents. 

28. 

40  and  5  %. 

12. 

$2351.25. 

65iff  cents 

9. 

10  bottles. 

29. 

60  and  10%. 

2. 

$  13,227.50. 

10. 

4  men. 

30. 

40  and  15  0/,. 

Page  631. 

3. 

104  days. 

13. 

3\V- 

4. 

$591.09+. 

Page  625. 

1. 

$135. 

14. 

1  mi.  95  rd. 

5. 

tV^t- 

3. 

12. 

2. 

10% 

1  yd.  6  in. 

6. 

$  782,300. 

4. 

8. 

3. 

$150. 

16. 

£23. 

7. 

$100. 

5. 

70. 

4. 

10%. 

17. 

$6000; 

8. 

$  1005.50. 

6. 

ism 

5. 

$  8.96. 

$  14,000. 

9. 

eoo/o. 

16 


J 

Page  634. 

Page  637. 

5. 

18.708+. 

10. 

$171.98; 

1.   $100.02. 

10.   5  acres  ; 

6. 

27.532-. 

$  194.01  ; 

2.   $500. 

23^  days. 

7. 

28.408-. 

$  174.08 ; 

3.    $1.75   dis- 

8. 

37.202+. 

$167.87. 

count  per 

1.    $327.05+. 

9. 

43.290-. 

$1000. 

2.    $154.44+. 

10. 

63.245-. 

Page  643. 

4.  $700. 

5.  $198.80. 

3.  $291.08-. 

4.  $1276.28+. 

1 

17. 
12. 

1. 

.316+. 

X. 

2. 

6.    33  days. 

5.   $874.75+. 

2. 

.632+. 

3. 

36. 

7.   Sight. 

3. 

.949-. 

4. 

29. 

8.    7%. 

Page  638. 

4. 

.316+. 

5. 

28. 

9.    Par. 

1.    $2109. 

5. 

.632+. 

6. 

33. 

10.  50/0. 

2.    $95.25. 

6. 

.949-. 

7. 

73. 

3.  $40. 

4.  $34.37|. 

7. 
8. 

1.265-. 
1.586+. 

8. 
9. 

48. 
16. 

1.    23  inches. 

2.    1  ft.  U  in. 

5.    $5106.25. 

9. 

1.897+. 

10. 

113. 

3.    120  rods ; 

6.    $1080;  4%. 

13. 

2.214-. 

11. 

180  yards. 

2640  yards. 

7.    100/0;  5fo/^. 

14. 

2.530-. 

12. 

10  rods. 

8.   3f%. 

15. 

2.846-. 

13. 

12rd.;18rd. 

9.   i%- 

16. 

3.162+. 

14. 

1^  acre. 

Page  635. 

17. 

.348-. 

4.    66  feet ; 

Page  639. 

18. 

.379+. 

Page  644. 

7.92  inches. 

10.    $62.50; 

19. 

.411+. 

15. 

$420. 

5.    15  inches. 

i%- 

20. 

.443-. 

16. 

192  rods. 

8.  12|  inches. 

9.  24  inches. 

17. 

50  rods ; 
1430  yd ; 

2.    Increased, 

3. 

16,203.03. 

10.   5  feet. 

$68.75. 

21^  acres. 

3.   $357.42. 

18. 

150  yards. 

4.   $175. 

Page  641. 

19. 

396  yards. 

Page  636. 

5.    6'8. 

4. 

172,030. 

20. 

113.14-  rd 

1.    $8.40; 

6.    East,  15°. 

5. 

9%. 

(18). 

7.   7  a.m. 

6. 

$1160.32+. 

1. 

161%- 

2.   $629.20. 

8.  $2437.60+. 

7. 

$592.48-. 

2. 

3  minutes. 

3.    $650.39-. 

9.  $4800; 

8. 

$120.76; 

3. 

9;*;  3^ 

4.   $225.16+. 

($4797.58-). 

$162.61; 

5.    $343.61-. 

$62.79; 

Page  645. 

($343.43+). 

Page  640. 

$83.72; 

4. 

300%. 

6.   $484.75; 

1.   2.646-. 

$93.38. 

5. 

7  X  10,  etc. 

($485). 

2.   3.742-. 

6. 

$1.50. 

7.    $60.63+. 

3.  6.164+. 

Page  642. 

7. 

100%. 

8.   $2. 

4.   8.602+. 

9. 

$  126.80. 

8. 

20. 

ANSWERS 

). 

17 

9.   12,000. 

(^)37i 

Page  650. 

Page  653. 

10.   48  cases. 

(e)  Impossi- 

9. 

$  1.72-. 

7. 

15.588  + 

ble. 

10. 

66f%. 

sq.  ft. 

1.    £45  16  s. 

11. 

11. 

8. 

1176  sq.  ft. 

2.    1  lb.  14  oz. 

12. 

10,000. 

9. 

672  sq.  ft. 

3.    10  men. 

Page  648. 

13. 

M;  H- 

10. 

3.464+  sq. 

4.    44  yards. 

4. 

$24. 

14. 

$  15,000. 

in. 

5.    3hr.l6min. 

5. 

1  mi.  7  fur. 

15. 

896  pounds. 

Diagonals,    2 

6.    319  rd.  4  yd. 

27  rd.  3 

16. 

472AV- 

in.  and 

1  ft.  6  in. 

yd.  2  ft. 

17. 

$  28.841 

3.464+  in. 

Hi  in. 

19. 

637i 

11. 

21.008+  feet. 

Page  646. 

6. 

L.C.M.  360 

20. 

$  166.581. 

12. 

50.2656  sq. 

7.    10  feet. 

7. 

194.45+  ft. 

21. 

6000  copies. 

in. 

8.    4  acres ;  16 

8. 

$6500. 

22. 

183-^  years. 

13. 

3.1416  a;2. 

acres. 

9. 

22f%. 

23. 

28ff|  bbl. 

14. 

10  inches. 

9.    2^  pounds. 

10. 

1200 

24. 

80  rods. 

15. 

19.635  sq.  ft. 

10.   432  pounds. 

peaches. 

11.   6  days. 

Page  654. 

12.   40/0. 

1. 

$4.45-. 

Page  651. 

1. 

$215.75. 

13.   4^o/o. 

2. 

$3.60+. 

2. 

$1091.66+. 

2. 

$414.33+. 

14.   41 0/0. 

3. 

128f%. 

3. 

$  125.90. 

15.  18000. 

4. 

$34,312.50. 

4. 

$  104.55. 

16.  $1000; 

Page  649. 

5. 

$  86.44-. 

$  1400. 

3. 

$5.76. 

6. 

$1551.27+; 

Page  655. 

4. 

$  4.38. 

($  1552.06-). 

5. 

$157.68+. 

Page  647. 

5. 

$1.17+. 

7. 

$6000. 

6. 

$453.61+. 

17.    $70,000. 

6. 

$16.72-. 

18.  68  rd.  3  yd. 

7. 

$11.94. 

Page  652. 

1. 

84  sq.  ft. 

4  in. 

8. 

$51.40. 

8. 

(a)  $  33,000  ; 

2. 

234  sq.  yd. 

19.    $165;  $210; 

9. 

$  76.80. 

(6)1310/,. 

3. 

264  sq.  rd. 

$225. 

10. 

$  92.38+. 

9. 

N.Y.  &N.  E. 

4. 

84  sq.  in. 

20.    $315. 

$50.50. 

5. 

990  sn   ft 

uov  hq.  ib. 

21.   Qt\\%%; 

1. 

$1000. 

10. 

40  acres. 

6. 

900  sq.  ft. 

6%. 

2. 

$189.92- 

7. 

420  sq.  yd. 

22.    6  hours. 

3. 

$  31,000. 

2. 

1470  sq.  ft. 

4. 

m%- 

3. 

294  sq.  rd. 

Page  656. 

1.    $3.37.68+. 

5. 

.7525  miles. 

4. 

1764  sq.  rd. 

8. 

330  sq.  rd. 

2-    1.5625. 

6. 

10mo.l7da. 

5. 

300  sq.  yd. 

9. 

744  sq.  rd. 

3.    (a)2j\%. 

nearly. 

6. 

42.332+  ft.; 

10. 

240  sq.  ch. 

(ft)  Vf. 

7. 

(lO^perbu.); 

2031.94- 

Cc)  180. 

H%- 

sq.  ft. 

1. 

2i5,ni. 

18 


ANSWERS. 


2.  1.975. 

3.  $100. 

4.  152.02. 

5.  $489.75, 

6.  93  lb.  9f  oz. 

7.  4ffi 

8.  $653.08-. 
10.    2,  5,  7,  13, 

23. 

Page  657. 

1.  3  months. 

Page  658. 

2.  9  months. 

3.  1  yr.  5  mo. 

4.  3|  months. 

Page  659^ 

5.  28^  days. 

6.  2^  months. 

7.  4^j  months. 

8.  7  mo.  26  da. 

9.  4^  months. 

10.  August  10. 

11.  94f  cent». 

12.  $31.50; 
$21; 
$31.50. 

13.  A,  $  3500 ; 
B,  $3600. 


Page  660. 

14.   60bu.;40bu. 

20  bushels. 

A, $36; 

B,  $24. 

A,  $875; 

B,  $  1458.33+; 

C,  $1166.67-. 
18.    A.  54  tons  : 


15. 
16. 

17. 


B,  31^  tons ; 

C,  94^  tons. 
19.    6f  quarts. 


1.  186,441. 

Page  661. 

2.  34,538,549f. 

3.  82,739ffi 


$540. 
220%. 
$  440. 
2|  hours. 
$  1440. 
1350  sq.  in. 


8. 


Page  662. 

7.   $1.29. 
Feb.  12, 
1809. 
$5  gain. 
2025. 


9. 
10. 


$9.52. 

1.464375. 

56.65-  feet. 

$21. 

A,  $2; 

B,$3; 

C,  $3.50; 

D,  $4.50. 


Page  663. 

8.  M;  a). 

9.  4725  1b. 


1.  9600  men. 

2.  $37.75-. 

3.  $9.84+. 

4.  $4000. 


2.16603. 
7.95^%. 
5  yr.  5  mo. 
20  da. 

$3800.47+. 
58  feet. 


Page  664. 


10. 
11. 
12. 
13. 
14. 
15. 


8 A  feet. 


8 

7% 


1  yr.  10  mo. 
$9956.86-. 

$  759.76. 


Page  665; 

1.  336AV. 

2.  n^j  rods. 
$  1533.75. 
$116.36-. 
7  bu  ;  5  bu. 

1  pk.  3  qt. 
10  acres. 
.66+. 
21  lb.  5  oz. 

18  pwt. 

m  gr. 
7%. 

$  16,000. 
113.14^  rd. 


3. 
4. 
5. 

8. 

9. 

10. 


11. 
12. 
13. 


Page  666. 

14.    if 

4096  cu.  ft. 
$331.86^. 
223|  sq.ft. 
$28.01+. 

20.  5\i  tons. 

21.  $40.50. 

22.  56.8+%. 
23.-.  $  27.84+. 


15. 
17. 
18. 
19. 


24.  16- cents. 

25.  HI;  iH; 

26.  240  rods. 

Page  667. 

1.  6.2832  .r. 

2.  3.1416x2. 

3.  .7854  .t2. 

4.  .07958x2. 

5.  1017.8784 

sq.  ft. 

6.  7  yards, 

7.  50  rods. 

8.2!, 

2 

9.    11,250  sq. 
rd. 

10.  2688  sq.  ra. 

11.  62.35+  sq. 

ft. 

12.  1469.69- 

sq,  yd, 

13.  7.958  sq.  ft. 

14.  960  sq.  rd. 

Page  668. 

15.  4800  sq.  yd. 

16.  541.27-  sq. 

rd. 

17.  3800  sq.  rd. 

18.  234  sq.  rd. 
19;    779,42+  sq, 

yd. 
20;    36  feet. 

21.  40  rods, 

22.  93.53+  sq. 

in, 

23.  113.10-  sq, 

in. 
244.^  50  sq.  ft. 


Page  669. 

1.    $6224.35. 

6. 

210. 

J.  £7 

1888,  $27,307. 

25.   62.832sq.in. 

2.    2  yr.  4  mo. 

7. 

1260. 

1889,  $  29.499. 

3|da. 

8. 

594. 

1890,  $  25.772. 

1.    18,990.59+. 

9. 

3366. 

1891,  $  25.680. 

3.    i%. 

Page  672. 

10. 

3060. 

1892,  $27,801. 

4.   ?  49.02. 

3.    $1441.94+. 

5.    |21. 

4.    $150;  $270. 

1. 

$88.73+; 

Page  681. 

5.    lOo/o. 

$86.27-. 

1. 

96  sq.  in. 

Page  670. 

6.    288  boards. 

2. 

$309.37i; 

2. 

72  sq.  in. 

6.    $220.54-; 

7.    49  rods. 

$  464.061 ; 

3. 

144  sq.  in. 

($  220.67-). 

8.    $1540. 

$513.56^. 

5. 

75.3984    sq. 

7.    $310.85+. 

9.    14  rods. 

3. 

$27xV; 

in. 

8.    £1,411,734 

10.    $50,000. 

$32i;$29i; 

7. 

294  sq.  in. ; 

6s.  Id. 

$41i 

6x\ 

9.    $21,050. 

1.    $69.85-. 

8. 

6  inches. 

10.    $1200. 

Page  679. 

9. 

216  sq.  in. 

Page  673. 

4. 

$4945.05+; 

1.   $1400. 

2.    $  12,500. 

$5934.07-; 

Page  682. 

2.    $8.63-. 

3.    3o/„. 

$4120.88-. 

10. 

128  sq.  in. 

3.   $1917. 

7.    20. 

5. 

$42.96-; 

11. 

10  feet. 

4.    284  days. 

10.    14  weeks. 

$34.80-; 

12. 

1200  sq.  ih  , 

5.   357. 

11.    21  men. 

$28.35+; 

20  in. 

6.    143. 

12.    240  miles. 

$46.39+. 

13. 

702  sq.  in. 

7.    .8. 

13.    $96. 

6. 

6  men. 

14. 

4.7124  sq.ft. 

8.    14  feet. 

14.    $15,000. 

7. 

31f  days. 

15. 

48  sq.  in. 

15.   $128. 

8. 

208  acres. 

17. 

64  sq.  in. 

Page  671. 

9. 

240  men. 

18. 

576  sq.  in. 

9.    3  mi.  207  rd. 

Page  677. 

10. 

20  days ; 

19. 

6.93+ sq.  in. 

1  yd.  1  ft. 

$  432.G3-. 

24  days. 

20. 

37.6992    sq. 

6  in. 

in. 

10.    33mi.225i\ 

Page  678. 

Page  680. 

21. 

122.5224  sq. 

rd. 

1.    306  sq.   ft.; 

11. 

$13. 

in. 

11.    1  hr.  2  min. 

12  ft. 

12. 

$226.80. 

52  sec.  P.M. 

2.    126  sq.  yd. ; 

13. 

1363^  lb. 

Page  683. 

12.   $72. 

12  yd. 

14. 

12i  days. 

22. 

13  inches. 

13.    .6  week. 

3.    1110  sq.  rd.; 

Table. 

23. 

122.5224  sq. 

14.    $34.26+. 

15  rd. 

1883,  $  25.923. 

in. 

15.  '421  yards. 

4.    210  sq.  ch.  ; 

1884,  $  26.254. 

16.    175  sheep. 

15  ch. 

1885,  $  28.972. 

1. 

130,548  ses- 

18.   $425.51. 

5.    600  sq.  in. ; 

1886,  $  26.306. 

sions. 

19.    llf  rolls. 

16  in. 

1887,  $27,543. 

2. 

11  poems. 

20 


ANSWERS. 


3.  1237.40; 
($  237.52). 

4.  $10,000. 


Page  684. 

5.  $40. 

6.  303  feet. 


Page  686. 

1.  $10.34+; 
$  505.38- 
$78,133^. 


Page  688. 


15. 


lO. 


57  cents. 

4  lots. 

$55.15-. 

$7600. 

285  acres  ; 

$213.75 

cornniis.sion. 

$425.52+; 

$443; 

($446.26+). 

$900; 

($899.54+). 
$1394.05. 


Page  687. 

11.  m  tons. 

12.  2l5  acrsR. 

13.  .56;  .0012. 

14.  144  yards. 

15.  26j^  cents. 

16.  l^^x  cents. 
$441. 
$12,204. 
27  sq.  yd. 


17. 
18. 
19. 
20. 


21. 


22. 
23. 


24. 
25. 

26. 
27. 
28. 
29. 


395,999.- 

922186 ; 
?H;  I- 

24|  yards. 

$  2022.22+  ; 

$371.44-; 

($371.62^). 

20%. 

$  1280.42+  ; 

($1279.54+). 

$  8,575,875. 

$54,368.52|. 

3960  inches. 

^1  %  gain  ; 


700  sq.  in.; 
896  sq.  in. ; 
1568  cu.  in, 


30.   67,750  acres. 


Page  689. 

1.  18  cu.  ft. 

2.  400  cu.  in. 

3.  60  cu.  in. 

4.  82^  cu.  yd. 

5.  83.14-     cu. 

ft. 

6.  30.56-    cu. 

ft. 

7.  169.65-  cu. 

meters. 

8.  1163.29- 

gal. 

9.  9719.325  lb. 
10.  57|  cu.  in. 


Page  690. 

12.  3^  inches. 

13.  301.59+  cu. 

yd. 

14.  13  inches. 


9. 


10. 


1.  $1,468,380,- 

830. 

2.  1,001,101. 

3.  202,100,001. 

00006. 

6.  $1983.38}. 

Page  691. 

7.  $2,382+. 

8.  lOT.  17c\vt. 

3  qr.  8  oz. 
$51,000; 
$1260; 
$49,740. 
$353,369,- 

654.14; 

94.84-  %. 

11.   $47,891,- 

78.^.50 ; 
3.9083+  %. 

1/5.  OgJJ      /(,. 

Page  692. 

13.  $8000. 

14.  Due.  $11,- 

646.19. 


Page  693. 

1.  16  (board)  ft. 

2.  7  (board)  ft. 

3.  8  (board)  ft. 

4.  14  (board)  ft. 

5.  4  (board)  ft. 
18  (board)  ft. 
7  (hoard)  ft. 
24  (board) ft. 
14  (board)  ft. 


10.  16  (board)  ft. 

11.  40J    (board) 

ft. 

12.  27  (board Jt. 

13.  30  (board)  ft. 

14.  15  (board)  ft. 

15.  9  (board)  ft. 

16.  $16.20. 

17.  1575 (board) 

ft. 

18.  $3.56. 


Page  694. 

19.  $60.48. 

20.  $52. 


m 

6  lots. 

$  5889f 
36  days. 
108. 

146.86+  mi. 
B,  $1000; 
C,  $1500. 


Page  695. 

9.   $5.94— 


1.  682.32  sq.ft. 

2.  1 17.81 sq.  in. 

3.  50  sq.  in. 

4.  28.54  sq.  in. 

5.  7.854  in ; 

7.071+  in. 

6.  78.54  sq.  in.; 

28.54  sq.  in. 

7.  3.1416  sq. 

in. ; 
12.5664  sq. 
in. 


8.    1  to  9 ;  area 

14. 

i\.i.y  o  vv 

ItV 

xuxvo 

8. 

185.61-   sq. 

(6)  .003672; 

=i22x3.1416. 

15. 

n- 

ft. 

(c)  1600. 

9. 

192  shares. 

16.   (a)f; 

Page  696. 

Page  701. 

10. 

$3200. 

(c)  .125  acre. 

9.    43.61  pq_  yd. 

1. 

113.0976  cu. 

11. 

3450  copies. 

17.   25%. 

iO.    392.7  sq.ft. 

in. 

12. 

A,  $375; 

18.   A,  $50; 

11.    84.8232  sq.  in. 

2. 

14.1372    cu. 

B,  $150; 

B,  $13,600; 

12.     3:1. 

in. 

C,  $100. 

C,  $4350. 

13.    500  sq.  ft. 

3. 

1:8. 

13. 

$  1059.35+ ; 

14.   500  sq.  ft. 

4. 

.5236:1. 

($1059.39-). 

Page  707. 

15.    1:4. 

19.    100  miles. 

Page  704. 

20.   $2000. 

Page  698. 

Page  702. 

14. 

$1434.29+. 

22.   (a)i; 

16.    12.5664  sq.  in. 

5. 

113.0976  in.; 

15. 

2  :  58  :  48 

(c)  M- 

17.   31  cents. 

4071.5136 

P.M. 

23.    (a)  If; 

18.   Equal. 

sq.  in. ; 

16. 

19i|  days. 

(b)  .00375 ; 

19.    2:3. 

24,429.0816 

17. 

71  days. 

(c)  .16. 

20.    127.328  sq.in. 

cu.  in. 

18. 

$  13,000. 

6. 

7. 

245  lb.  7  oz. 
2:3. 

19. 

2013.7824 
sq.  in. 

Page  708. 

25.   (6)4000; 

1.    11%;  |90. 

2.   $88,922.4231. 

8. 

.4764 ; 

20. 

452.3904  cu. 

(c)  13.163; 

about  f 

in. 

(d)  1.706. 

Page  699. 

9. 

.2146; 

27.   $331 

3.    !?  13,173.60. 

about  |. 

2. 

(«)VA; 

28.   $200,000. 

4.   $1121 

(&)tV 

29     2  men 

f<J%J  m           4mJ      iilOll. 

5.   $6050. 

1. 

44  men. 

30.  $75. 

2. 

A,  $100; 

Page  705. 

Page  700. 

B,  $120; 

6. 

$17,728.53-. 

Page  709. 

1.    13. 

C,  $  120. 

7. 

15  hours. 

32.   (a)  7; 

2.   21. 

3. 

7.62+ ft.; 

8. 

$  73.80. 

{c)  m 

3.   32. 

(16.89+  ft.). 

9. 

4  mi.  per  hr. 

33.  (a)  tV  ; 

4.   41. 

4. 

$821.76+; 

10. 

10  hours. 

ib)  22  rd.  4 

5.   53. 

($  825.50). 

yd.    2   ft. 

6.   62. 

Page  703. 

Page  706. 

If  in.  ; 

7.   75. 

5. 

$4160.30-; 

12. 

(a)  8; 

(c)  .0625. 

8.   82. 

($4157.60-). 

(^)i|; 

35.   (6)20,007.- 

9.vf- 

6. 

452.3904    sq. 

(c)  .625. 

253; 

10.  H- 

in.; 

13. 

(a)2rd.lyd. 

(c)  .00003. 

11.  M- 

904.7808  cu.  in 

14. 

(a)  .001, 

36.   (a)  .00091; 

12.   1.5. 

7. 

3  hr.  13  min. 

.0001024, 

(6)  00006. 

13.   If 

36  sec.  fast. 

32.004  ; 

37.   8  men. 

^: 


2: 

ANSWERS. 

38.  $1333^; 

3.    122°  26^ 

4.   A,  $105; 

4.  23,048,771 

$2000; 

15^^  West 

B,  $87.50. 

sq.  in. 

$2666|. 

4.   .00005. 

5  miles. 

5.   Thos.,  $2.25; 

5.   A,  $2870; 

5.   $5600; 

Henry,    $1.35; 

Page  710. 

B,  $7175; 

$97,920. 

Richard,  $0.54. 

39.   fl514. 

C,  $  1435. 

6.    233|  qt.  dry 

40.   $60. 

6.  $545.82+. 

Page  716. 

measure. 



7.  8o/o. 

6.   30,013; 

7.  Iff  minute. 

1.    113. 

8.    117  feet. 

.1716-. 

2.   124. 

9.  2.16+. 

7.  26  days. 

Page  719. 

3.   155. 

10.    16|o/o. 

8.  $40.80. 

8.    A,  56  times  ; 

4.   341. 

9.   $189;  $147. 

B,  35  times ; 

5.   2.35. 

Page  714. 

10.   17.43-%. 

C,  22  times. 

fi     4  Oft 

1.  $44.83i. 

2.  IE-. 

11      7  mo    6  fin 

o.    ^.v/O. 

±X.       i    IIIO.    O  Udi. 

12.   The  latter. 

1.  $756.96. 

Page  711. 

3.  $25.62^. 

13.    gVless. 

2.  $1530. 

1.    $109; 

4.  $9.80. 

3.  $821.57+. 

$  15.95-. 

5.   18  days. 

Page  717. 

4.  $1178.46. 

2.  Sp/,; 

6.  45  men. 

14.  $5145. 

15  years. 

7.  $44,092. 

15.   5  miles. 

Page  721. 

3.    $600; 

8.  $65. 

16.    131^  miles. 

1.  $150. 

90  (87)  days. 

9.  $246.36+; 

2.  $360. 

4.   January  12. 

($246.15+). 

1.   .023825+; 

3.  5  miles. 

6.   $950; 

18|%; 

4.   162  miles. 

$5937.50. 

UU- 

5.   15,840  feet. 

6.   2^0/^; 

Page  715. 

2.  $715. 

6.    760  acres. 

m%- 

10.  £6  3  8. 

3.   18.000002+; 

7.    60  feet; 

IHd- 

7.745967-. 

Page  722. 

231. 
8.   2Heet; 

1        *W*)  U~i  fVonr>a 

1.    17,000; 

Page  718. 

J..      K'fTt'.t'O  ir<lI10s. 

2.    22.525  sq.  m. 

263.8944. 

.0002938. 

4.    2945  kilos 

3.    176.7158q.m. 

9.   171.65+ sq. 

Ist,  great- 

nearly. 

4.    93.15  ares. 

in. 

est; 

5.   $1521.30- 

5.    1029  ares. 

2d,  least. 

($1520.27+); 

Page  712. 

.000003125. 

$1837.90- 

Page  723. 

10.    351.8592  sq. 

2.  /^V; 

($  1836.73  f). 

6.    276.25  francs. 

in.  ; 

5t^^; 

7.   5 16,000 liters. 

502.656    cu. 

t\,  n- 

1.    Latter,  ^ 

8.   296  bottles. 

in. 

3.   £.4545-; 

greater. 

9.   306  marks. 

28,475  m.  ■ 
$35.17i. 

2.   .909. 

10.   55,200  kilos. 

2.  3|. 

3.  $101.86+. 

11.   .62137^  mi. 

ANSWERS. 

23 

12.  15,748  feet. 

9.   192f  bushels; 

13. 

35. 

17. 

13  years. 

1  ^     fi4  rn    in  • 

231  cu   in 

14. 

9. 

XO.      O^   LU.    111.  , 

10.   99  yards. 

15. 

-^• 

1. 

10  cents. 

14.   2f|  bushels; 

16. 

2. 

27iffgal. 

Page  729. 

17. 

2. 

3.   -Qxy. 

18. 

4. 

Page  737. 

4.    11  a6c. 

19. 

6f. 

2. 

8. 

Page  724. 

5.   -'Sxyz. 

20. 

120. 

3. 

50  cents. 

15.   2^1- pounds. 

6.   x  +  Q. 

21. 

120. 

4. 

2  pigs. 

16.  25,2525.25+ 

7.   _10a-8.T. 

22. 

186f. 

5. 

40  years. 

miles. 

8.   -a+5  6-4c. 

23. 

12. 

6. 

4  cents. 

17.  Hfsq.yd. 

9.   l^a;-l. 

24. 

h 

7. 

12  yards. 

18.   2ffff  acres. 

10.   3ia;  +  38i-. 

25. 

12. 

8. 

24  cents. 

19.   20f  f  rods. 

9. 

6. 

20.   STaV  c«-  ft. 

Page  731. 

Page  735. 

- 

21.   16|^  grains. 

9.    2a;  +  4. 

1. 

15. 

1. 

4. 

22.   1.584  Km. 

10.   8a;  +  l. 

2. 

96. 

2. 

5. 

11.   -4  a; +  9. 

3. 

20  years. 

3. 

12. 

12.   -2x4-20. 

4. 

11600.' 

4. 

7. 

Page  725. 

13.   4  a; +  5. 

5. 

96  marbles. 

5. 

24. 

23.  1 19.82-. 

14.   -x-3. 

6. 

Father,    88 

24.   289.56- mi. 

15.  2x  +  3a  +  4. 

years ; 

Page  738. 

25.   1 24.80-. 

16.   15.V-2-5  6. 

son,  50  years 

.    6. 

4. 

17.  -2d  +  e+f: 

7. 

20  cents. 

7. 

6. 

Page  726. 

8. 

3  years. 

8. 

9. 

1.  101  lb.  7  oz. 

Page  732. 

9. 

81. 

12  pwt.  12 

1.    6. 

10. 

50. 

gr. 

2.   9. 

Page  736. 

11. 

Coat,  $  12 ; 

3.   33  mi.  172  rd. 

3.   6. 

9. 

48  gallons. 

vest,  1 3. 

2  ft.  If  in. 

4.   4. 

10. 

432. 

12. 

3. 

5.   9. 

11. 

118;  548. 

13. 

8. 

Page  727. 

12. 

14   two-dol- 

14. 

x=1,y  = 

=  8. 

4.    1  hr.  13  min. 

Page  733. 

lar  bills; 

15. 

a;  =  9,  3/  = 

=  6. 

36f  sec. ; 

7.  9. 

15   five-dol- 

16. 

a;  =  3,  y  = 

=  4. 

2Q°  15'  east. 

8.   6. 

lar  bills. 

17. 

a:  =  5,  y= 

10. 

5.  1 96  gain. 

9.  6. 

13. 

6  years  ; 

6.   18.8496  feet; 

10.   3f 

36  years. 

Page  739. 

28.2744  sq.  ft. 

; 

14. 

12;    15. 

18. 

a;  =  6,  y  = 

=  8. 

12.5664  cu.in. 

Page  734. 

15. 

$3,  son ; 

19. 

x  =  \,y  = 

=  1. 

7.   48^;  $1.20. 

11.   11. 

$4,  father. 

20. 

^=H.  2/= 

^^ 

8.  40  acres. 

12.   11. 

16. 

32;  15. 

21. 

x-ll,y  = 

=7. 

24 


ANSWERS. 


22.  a;  =  2,  y  =  3. 

23.  x=l2,y  =  3. 

24.  a:  =  l,  y  =  2. 

25.  a;  =  3,  y  =  4. 

26.  a:  =  iy=i 

27.  .'c  =  7, 3/  =  5. 

28.  tc  =  4,  y  =  -  4. 

29.  a;  =  2,  y  =  19. 

Page  740. 

30.  a;  =  4,  y  =  24. 

31.  a;  =  42,j5/  =  63. 

32.  x  =  5,y  =  7. 

33.  a;  =  64,000, 

y  =  36,000. 

34.  a;  =  6, 3/  =  3. 

35.  x  =  8,y  =  9. 

36.  a;  =  ll,y  =  15. 

37.  a;  =  23,y  =  18. 

38.  a;  =  13,  y  =  9. 

39.  x=10,  y  =  5. 


10. 

11. 
12. 
13. 
14. 


8. 


9. 


Raisins,  12  jf' 
cheese,  17  ^. 
12  and  7. 
If. 
iff- 
40  pounds. 


1.  15  and  22. 

2.  47  and  19. 

3.  5  and  4. 

4.  23  and  42. 

Page  741. 

5.  19    two-dollar 

bills ; 
13  five-dollar 
bills. 

6.  10  pigs;  15 

sheep. 

7.  Oranges,  3  f; 
peaches,  2  f. 
Tea,  60  ^ ; 
cofiFee,  25  ,^. 
4  horses,  16 

cows,  32  sheep, 
2  pigs. 


Page  742. 

15.   40  pounds 
green  tea  ; 
60  pounds 
black  tea. 

Page  743. 

2.  x  =  2,  y  =  3, 

2=5. 

3.  a;  =  7,y  =  13, 

2=1. 

4.  a;  =  12,y  =  31, 

2  =  19. 

5.  a;  =  10,y  =  13, 

2=16. 

Page  744. 

6.  a:=12,y  =  18. 

7.  «=12,  y  =  6. 

8.  a;  =  6,  y  =  5. 

9.  a:  =  19^, 
y  =  -17. 


1.  $30,000. 

2.  A,  20  chest- 

nuts; B,  2 
chestnuts. 

3.  |5;  $3. 

4.  17,38,  and  45. 

Page  745. 

5.  21  and  32. 

6.  1 18;  |32; 

1 16. 


7.  $600. 

8.  $150. 

9.  66^  acres ; 
38f  acres ; 
25y\  acres. 

10.  70  yards. 

11.  $20. 

Page  746. 

1.  a;2  +  7a;  +  10. 

2.  a;2  +  17a;  +  72. 

3.  2a;2  +  9a;-|-10. 

4.  23^2  +  26.1; 

+  72. 

5.  3x2  +  22x 

+  7. 

6.  4  a;'  -f  4  a;  +  1. 

7.  a;' +  6  a; -27. 

8.  a;2-f-a;-42. 

9.  a;2-25. 

Page  748. 

1.  ±7. 

2.  ±5. 

3.  ±5. 

4.  ±5. 

5.  ±5. 

6.  24. 

7.  24. 

8.  ±6. 

9.  13. 

10.  5. 

11.  .-t7. 

12.  ±1. 

13.  ±8. 

14.  ±4. 

15.  ±6. 

16.  7. 

17.  13. 

18.  ±6. 

19.  13. 


20.  13. 


1.  60  rods;  30 

rods. 

2.  4  inches. 

3.  10  and  8 

4.  45. 

5.  50. 


7. 


Page  749. 

6.   2500. 

12  rods;  20 

rods. 
12  yards. 
8  feet. 
24  and  25. 


Page  751. 

1.  a; +  3  =±7. 

2.  a; -6  =±8. 

X  -  4  =±  6. 

x-8=dk5. 

x  +  9=±  10. 

X  -f  1  =±  5. 

a;- 7  =±8. 

a;-ll=±12. 

a: +  7  =±10. 

a;-ll=±13. 


3. 
4. 
5. 
6. 
7. 
8. 
9. 
lO 


3. 

4. 

6. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 


7or-l. 

18  or -6. 

6  or -8. 

5  or  -  23. 

13  or  1. 

10. 

5  or  -  25. 

2  or  -  28. 

24. 

24  or  - 16. 

3or  1. 

5  or -35. 


ANSWERS. 

2, 

13.   1  or  -  29. 

11.   ^(7=  42  ft. 

3.    35°;      40°; 

84.8232  sq. 

14.   4  or -26. 

^Z>=24ft. 

105°;    105°. 

in. 

15.   8. 

16.   2  or -38. 

Page  784. 

Page  793. 

1.    72  cu.  in. 

9.     1:4. 

10.    14.1372    sq. 

2.   36  cu. in. 

Page  752. 

10.   90°,  5  in.  ; 

in. 

3,   36  cu.  in. 

1.   3  or -4. 

liin.,2in., 

11.    1|  in.,  3  in.; 

2.   5  or -2. 

2i  in. ;  37°, 

1  in.,  3  in. 

Page  798. 

3.    -  1  or  -  4. 

53°,  90°. 

4.    36  cu.  in.  ; 

4.    8  or  -  1. 

Page  794. 

124.71- cu.  in. 

5.    -4  or -5. 

Page  785. 

12.    108°. 

5.   690  cu.  in. 

6.    7  or  4. 

11.   ft  in. 

14.    Two,  15  in.; 

6.   62.882  cu.  in. 

7.  _6or-7. 

8.  19  or -4. 

two,  13  in.  ; 
384  sq.  in. 

1.   60  feet. 

Page  799. 

9.    18or-l. 

2.   48  feet. 

15.    240  sq.  in. 

7.    144  cu.  in.  ; 

10.   -  18  or  -  1. 

17.   40  sq.  ft. 

18  cu.  in. 

Page  786. 

8.    126cu.  in. 

1.   3,-2. 

3.    109^3- feet. 

Page  795. 

9.   1350  cu.  in.; 

2.   4,  -  5. 

4.   2160  feet; 

20.    122.5224  sq. 

50  cu.  in. 

3.   3,  -  7. 

2060  feet. 

in. 

10.    1300cu.in. ; 

4.   8,  2. 

5.    124  feet. 

21.    175  sq.  in. 

520  sq.  in. ; 

5.   9,  -  7. 

22.   257.6112  sq. 

in.     770  sq.  in. 

6.   3,  -  6. 

Page  787. 

23.    27  feet. 

7.   5,4. 

6.    13y%  chains. 

Page  800. 

8.    1,  -8. 

7.    162  feet. 

Page  796. 

11.    1300  cu.  in. 

9.   2,-9. 

8.   84  feet. 

24.    452.3904  sq. 

12.    65icu.  ft. 

10.   4,  1. 

ft. 

13.   2232  cu.  in. 

Page  788. 

254.4696  sq. 

ft. 

Page  753. 

9.    108  feet 

25.   4166|  mi. 

Page  801. 

1.   4  and  8. 

10.    119  yards. 

26.   2000  miles. 

14.    1927.3716 

2.  80  feet. 

3.  60  yards. 

27.  12,500  miles. 

28.  1:2. 

• 

1.   2.0944  in.; 

cu.  in. 
15.   929.9136  cu. 

4.   25  yards. 

4.1888  in.; 

29.   2828.4+  mi. 

in.  ; 

5.   5  feet. 

6.2832  in.; 

4.0256-  gal. 

6.   4  feet. 

8.3776  in. ; 

Page  797. 

7.  68  rods. 

10.4720  in. 

30.    113.0976  sq. 

Page  803. 

2.   2  inches ; 

in. 

16.   3053.6352  cu. 

Page  754. 

3.464+  in. ; 

31.   56.5488  sq. 

in. 

8.   34  feet. 

4  inches ; 

in. ; 

17.  381. 7044  cu.  in 

9.   65  feet. 

3.464+    in.; 

28.2744  sq. 

190.8522  cu. 

10.   3ifeet. 

2  inches. 

/ 

0F  THJC 

in. 

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